Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in PyPy3 (7.3.0) (judge verdict: Wrong Answer)

n = int(input())
start = 0
end = 0
left = 0
double = []
right = 0
for i in range(n):
    s = str(input())
    kienaileft = 0
    kienairight = 0
    for j in range(len(s)):
        if s[j] == ")":
            kienaileft += 1
        else:
            break
    for j in range(len(s)):
        if s[-(j+1)] == "(":
            kienairight += 1
        else:
            break
    nowleft = 0
    nowright = 0
    for j in range(kienaileft,len(s)-kienairight):
        if s[j] == ")":
            nowleft += 1
        else:
            nowright += 1
    num = min(nowleft,nowright)
    kienaileft += nowleft -num
    kienairight += nowright -num
    if kienaileft != 0:
        if kienairight != 0:
            double.append([kienaileft,kienairight])
        else:
            left += kienaileft
    else:
        if kienairight != 0:
            right += kienairight
  # print(left,double,right)
if left + sum([double[i][0] for i in range(len(double))]) != sum([double[i][1] for i in range(len(double))]) + right:
    print("No")
    exit()
if len(double) != 0:
    if not left or not right:
        print("No")
        exit()
else:
    if left == right:
        print("Yes")
        exit()
double.sort(key=lambda x:x[1])
double.sort()
double.sort(key=lambda x:x[0]-x[1])
now = left
for a,b in double:
    if now < a:
        print("No")
        exit()
    now += (b-a)
print("Yes")