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;

int main(){
  int N;
  cin >> N;
  
  vector<pair<int, int>> c(N);
  for(int i = 0; i < N; i++){
    string S;
    cin >> S;
    c[i].first = 0;
    c[i].second = 0;
    for(int j = 0; j < (int)S.size(); j++){
      if(S[j] == '(') ++c[i].second;
      else --c[i].second;
      c[i].first = min(c[i].second, c[i].first);
    }
  }
  sort(c.begin(), c.end(), greater<pair<int, int>>());
  
  int sum = 0;
  bool able = true;
  for(int i = 0; i < N; i++){
    if(sum + c[i].first < 0){
      able = false;
      break;
    }
    sum += c[i].second;
  }
  
  if(able && sum == 0) puts("Yes");
  else puts("No");
  
  return 0;
}