Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1

return 'Yes' if check(S1) or check(S2) else 'No'

            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1

            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

                val = val - 1
            if val < 0:

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

                val = val - 1
            if val < 0:
                return False

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

return 'Yes' if check(S1) or check(S2) else 'No'

                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

            if val < 0:
                return False

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

            if val < 0:
                return False
        return True if val == 0 else False

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

return 'Yes' if check(S1) or check(S2) else 'No'

            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:

        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

                return False
        return True if val == 0 else False

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:

return 'Yes' if check(S1) or check(S2) else 'No'

                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:

                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

        return True if val == 0 else False

    S1 = lst[0] + lst[1]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False

return 'Yes' if check(S1) or check(S2) else 'No'

        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False

        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]

return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]

    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]

return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """

        return []
    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """

    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """

    if k == 0:
        return []
    arr.sort()


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """

    if k == 0:
        return []
    arr.sort()
    ans = arr[-k:]


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """

    if k == 0:
        return []
    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:

    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:

        return []
    arr.sort()
    ans = arr[-k:]


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:

        return []
    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []

    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []
    arr.sort()


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []
    arr.sort()


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:

        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []
    arr.sort()
    ans = arr[-k:]


Prompt:
def solution(lst):
    """Given a non-empty list of integers, return the sum of all of the odd elements that are in even positions.

    Examples
    solution([5, 8, 7, 1]) ==> 12
    solution([3, 3, 3, 3, 3]) ==> 9
    solution([30, 13, 24, 321]) ==>0
    """

return sum([x for idx, x in enumerate(lst) if idx%2==0 and x%2==1])

def check(candidate):

    # Check some simple cases
    assert candidate([5, 8, 7, 1])    == 12
    assert candidate([3, 3, 3, 3, 3]) == 9
    assert candidate([30, 13, 24, 321]) == 0
    assert candidate([5, 9]) == 5
    assert candidate([2, 4, 8]) == 0
    assert candidate([30, 13, 23, 32]) == 23
    assert candidate([3, 13, 2, 9]) == 3

    # Check some edge cases that are easy to work out by hand.


Prompt:
def add_elements(arr, k):
    """
    Given a non-empty array of integers arr and an integer k, return
    the sum of the elements with at most two digits from the first k elements of arr.

    Example:

        Input: arr = [111,21,3,4000,5,6,7,8,9], k = 4
        Output: 24 # sum of 21 + 3

    Constraints:
        1. 1 <= len(arr) <= 100
        2. 1 <= k <= len(arr)
    """

return sum(elem for elem in arr[:k] if len(str(elem)) <= 2)

def check(candidate):

    # Check some simple cases
    assert candidate([1,-2,-3,41,57,76,87,88,99], 3) == -4
    assert candidate([111,121,3,4000,5,6], 2) == 0
    assert candidate([11,21,3,90,5,6,7,8,9], 4) == 125
    assert candidate([111,21,3,4000,5,6,7,8,9], 4) == 24, "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate([1], 1) == 1, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

            odd_collatz.append(int(n))

    return sorted(odd_collatz)

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

            odd_collatz.append(int(n))

    return sorted(odd_collatz)

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:

        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    if n%2==0:
        odd_collatz = []

        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"