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>
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000007
using namespace std;
int main(){
  int n;
  cin >> n;
  string s[n];
  vector<pair<int,int>> ls,rs;
  int sum=0;
  rep(i,n){
    cin >> s[i];
    int tmp=0,min=0;
    rep(j,(int)s[i].size()){
      if(s[i][j]=='('){
        tmp++;
        sum++;
      }else{
        tmp--;
        if(tmp<min)min=tmp;
        sum--;
      }
    }
    if(tmp>0)ls.emplace_back(min,tmp);
    else rs.emplace_back(min,tmp);
  }
  if(sum!=0){
    cout << "No\n";
    return 0;
  }
  sort(ls.rbegin(), ls.rend());
  sort(rs.rbegin(), rs.rend()); 
  int h=0;
  for(auto l: ls){
    if(h+l.first<0){
      cout << "No\n";
      return 0;
    }
    h += l.second;
  }
  for(auto r: rs){
    if(h+r.first<0){
      cout << "No\n";
      return 0;
    }
    h += r.second;
  }
  cout << "Yes\n";

  return 0;
}