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>> v(n);
  for (auto &e : v) {
    string s;
    cin >> s;
    int l = 0, r = 0;
    for (auto c : s) {
      if (c == '(') {
        l++;
      } else if (c == ')') {
        if (l > 0) {
          l--;
        } else {
          r++;
        }
      }
    }
    e = {l, r};
  }

  sort(v.begin(), v.end(), [](pair<int, int> a, pair<int, int> b) {
    return a.second == b.second ? a.first > b.first : a.second < b.second;
  });
  bool valid = true;
  int l = 0;
  for (auto e : v) {
    // cerr << e.first << ' ' << e.second << endl;
    if (l < e.second) {
      valid = false;
      break;
    }
    l -= e.second;
    l += e.first;
  }
  if (l != 0) valid = false;
  cout << (valid ? "Yes" : "No") << endl;
}