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;
using ll=long long;
#define MAX 500009
#define MOD 998244353
#define INF 1000000000

int main(){
  int N;
  cin>>N;
  vector<string> S(N);
  for(int i=0;i<N;i++){
    cin>>S.at(i);
  }
  vector<int> a(N,0),m(N,0),M(N,0);
  for(int i=0;i<N;i++){
    int n=S.at(i).size();
    for(int j=0;j<n;j++){
      if(S.at(i).at(j)=='('){
        a.at(i)++;
        M.at(i)=max(a.at(i),M.at(i));
      }else{
        a.at(i)--;
        m.at(i)=min(a.at(i),m.at(i));
      }
    }
  }

  int ans1=0,ans2=0;
  vector<pair<int,int>> ma,Ma;
  set<int> s;
  for(int i=0;i<N;i++){
    if(m.at(i)==0){
      ans1+=a.at(i);
    }else if(m.at(i)==a.at(i)){
      ans2-=a.at(i);
    }else if(a.at(i)>0){
      ma.push_back(make_pair(m.at(i),a.at(i)));
    }else if(a.at(i)<0){
      Ma.push_back(make_pair(m.at(i)-a.at(i),-a.at(i)));
    }else{
      s.insert(i);
    }
  }
  sort(ma.begin(),ma.end());
  reverse(ma.begin(),ma.end());
  sort(Ma.begin(),Ma.end());
  reverse(Ma.begin(),Ma.end());

  bool flag=true;
  int n=ma.size();
  for(int i=0;i<n;i++){
    if(ans1+ma.at(i).first<0){
      flag=false;
    }
    ans1+=ma.at(i).second;
  }
  n=Ma.size();
  for(int i=0;i<n;i++){
    if(ans2+Ma.at(i).second<0){
      flag=false;
    }
    ans2+=Ma.at(i).second;
  }
  n=s.size();
  for(int i=0;i<n;i++){
    if(ans1+m.at(i)<0){
      flag=false;
    }
  }
  if(flag&&ans1==ans2){
    cout<<"Yes"<<endl;
  }else{
    cout<<"No"<<endl;
  }
}