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 s[n];
    for(int i=0; i<n; i++) cin >> s[i];
    vector<pair<int,int>> t;
    int right = 0;
    int left = 0;
    for(int i=0; i<n; i++){
        int l=0,r=0;
        for(int j=0; j<s[i].size(); j++){
            if(s[i][j]=='(') l++;
            else if(l>0) l--;
            else r++;
        }
        if(r!=0 && l!=0) t.push_back(make_pair(l,r));
        else if(l==0) right += r;
        else{
            left +=l;
        }
    }
    sort(t.begin(),t.end(),[](pair<int,int> a, pair<int,int> b){
            return (a.first-a.second > b.first-b.second);});
    for(int i=0; i<t.size(); i++){
        cout << t[i].first << ' ' << t[i].second << endl;
    }
    for(int i=0; i<t.size(); i++){
        left -= t[i].second;
        if(left < 0){
            cout << "No" << endl;
            return 0;
        }
        left += t[i].first;
    }
    if(left == right) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}