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;
  string ss[n];
  for (int i=0; i<n; i++) cin>>ss[i];
  vector<int> left, middle, right;
  for (string s: ss) {
    int cnt = 0;
    bool not_left = false;
    int cnt2 = 0;
    for (char c: s) {
      if (c == '(') {
        ++cnt;
        ++cnt2;
      } else {
        --cnt;
        if (cnt2 > 0) {
          --cnt2;
        }
        if (cnt < 0) {
          not_left = true;
        }
      }
    }
    if (cnt > 0) {
      if (not_left) {
        middle.push_back(cnt);
      } else {
        left.push_back(cnt);
      }
    } else if (cnt < 0) {
      if (cnt2 > 0) {
        middle.push_back(cnt);
      } else {
        right.push_back(cnt);
      }
    } else {
      if (not_left) {
        middle.push_back(cnt);
      }
    }
  }
  long long left_sum = 0, right_sum = 0, middle_sum = 0;
  for (int i: left) {
    left_sum += i;
  }
  for (int i: middle) {
    middle_sum += i;
  }
  for (int i: right) {
    right_sum += i;
  }
  if (left_sum + middle_sum + right_sum == 0) {
    if (left.size() == 0 || right.size() == 0) {
      if (middle.size() == 0) {
        cout<<"Yes"<<endl;
      } else {
        cout<<"No"<<endl;
      }
    } else {
      cout<<"Yes"<<endl;
    }
  } else {
    cout<<"No"<<endl;
  }
}