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 Python (3.8.2) (judge verdict: Runtime Error)

#from collections import defaultdict, deque
#import itertools
#import numpy as np
#import re

def main():
    N = int(input())
    SS = []
    for _ in range(N):
        SS.append(input())
    S = []
    for s in SS:
        while '()' in s:
            s = s.replace('()', '')
        S.append(s)
    #print(S)
    S = [s for s in S if s != '']
    sum_op = 0
    sum_cl = 0
    S_both_op = []
    S_both_cl = []
    for s in S:
        if not ')' in s:
            sum_op += len(s)
        elif not '(' in s:
            sum_cl += len(s)
        else:
            pos = s.find('(')
            if pos <= len(s) - pos:
                S_both_op.append((pos, len(s)-pos))  #closeのほうが少ない、'))((('など -> (2,3)
            else:
                S_both_cl.append((pos, len(s)-pos))  #closeのほうが多い、')))(('など -> (3,2)

    # S_both_opは、耐えられる中でより伸ばす順にしたほうがいい？
    sort(S_both_op, key=lambda x: x[0])  #closeの数が小さい順にsort
    sort(S_both_cl, key=lambda x: x[0])  #これもcloseの数が小さい順にsort

    for p in S_both_op:
        sum_op -= p[0]
        if(sum_op < 0 ):
            print('No')
            exit()
        sum_op += p[1]
    for p in S_both_cl:
        sum_op -= p[0]
        if(sum_op < 0 ):
            print('No')
            exit()
        sum_op += p[1]
    
    if sum_op == sum_cl:
        print('Yes')
    else:
        print('No')


if __name__ == "__main__":
    main()