Problem: This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.

Solution (in PYTHON3):

Solution:
t = int(input())


def get_zero(a):
    prefixes = ''
    for idx in range(len(a) - 1):
        if a[idx] != a[idx + 1]:
            prefixes += str(idx + 1) + ' '

    if a[-1] == '1':
        prefixes += str(len(a))

    # print('prefixes: ', prefixes)
    return prefixes.strip()


for _ in range(t):
    n = int(input())  # String len
    a = input()
    b = input()

    c = 0

    prefixes = []
    for idx in range(len(a) - 1):
        if a[idx] != a[idx + 1]:
            prefixes.append(str(idx + 1))
            c += 1

    if a[-1] == '1':
        prefixes.append(str(n))
        c += 1

    rev_prefixes = []
    for idx in range(len(b) - 1):
        if b[idx] != b[idx + 1]:
            rev_prefixes.append(str(idx + 1))
            c += 1

    if b[-1] == '1':
        rev_prefixes.append(str(n))
        c += 1

    rev_prefixes.reverse()

    out = ' '.join(str(x) for x in prefixes) + ' ' + ' '.join(str(x)
                                                              for x in rev_prefixes)

    print("{} {}".format(c, out))

    # Output
    # k is number of operations used
    # list of prefixes

# Get to zero: 001011
# Second last
# 2 - 3
# 2
# 00 1011 -> 11 1011
# 3 - 4
# 3
# 111 011 -> 000 011
# 4
# 111111
#


# first discrepancy set:
# 2
# 001011 ->2-> 111011
# second descrepancy: 4
# 4
# 111011 -> 100011
# 1
# 000011
# 4
# 111111
# n
# 000000