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 fst(t) std::get<0>(t)
#define snd(t) std::get<1>(t)
#define thd(t) std::get<2>(t)
#define unless(p) if(!(p))
#define until(p) while(!(p))

using ll = std::int64_t;
using P = std::tuple<int,int>;

int N;
std::string S[1000100];
int I[1000100], m[1000100];
std::string T;

bool isValid(std::string &s){
    int cur = 0;

    for(char c : s){
        if(c == '('){
            cur += 1;
        }else{
            cur -= 1;
            if(cur < 0){
                return false;
            }
        }
    }

    return cur == 0;
}

int main(){
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);

  std::cin >> N;

  for(int i=0;i<N;++i){
      std::cin >> S[i];
  }
  
  for(int i=0;i<N;++i){
      int cur = 0, mn = 0;

      for(char c : S[i]){
          if(c == '('){
              cur += 1;
          }else{
              cur -= 1;
              mn = std::min(mn, cur);
          }
      }

      m[i] = mn;
  }

  std::iota(I, I + N, 0);
  std::sort(I, I + N, [&](int l, int r){return m[l] > m[r];});

  for(int i=0;i<N;++i){
      T += S[I[i]];
  }

  if(isValid(T)){
      std::cout << "Yes" << std::endl;
  }else{
      std::cout << "No" << std::endl;
  }
}