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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
constexpr ll mod = 1e9 + 7;
#define endl '\n'
 
int main() {
  ios::sync_with_stdio(0); cin.tie(0);
 
  ll N;
  cin >> N;
  string s;
  int total_all_l = 0, total_all_r = 0;
  for (int ii = 0; ii < N; ++ii) {
    cin >> s;
    int n = s.length();
    int open = 0;
    int all_l = 0, all_r = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i] == '(') {
        ++open;
      } else {
        if (--open < 0) {
          open = 0;
          if (total_all_l) {
            --total_all_l;
          } else {
            ++all_r;
          }
        }
      }
    }
    int close = 0;
    for (int i = n-1; i >= 0; --i) {
      if (s[i] == ')') {
        ++close;
      } else {
        if (--close < 0) {
          close = 0;
          if (total_all_r) {
            --total_all_r;
          } else {
            ++all_l;
          }
        }
      }
    }
    total_all_r += all_r;
    total_all_l += all_l;
  }
  
  if (total_all_l == 0 && total_all_r==0) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }


}