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> >v1, v2, v3, v4;
    for(int i = 1; i <= n; i++){
        string s;
        cin >> s;
        int x = 0, y = 0, mnbal = 0;
        for(int i = 0; i < s.size(); i++){
            if(s[i] == '('){
                x++;
            }
            else{
                y++;
            }
            mnbal = min(mnbal, x - y);
        }
        int bal = x - y;
        if(bal >= 0 && mnbal == 0){
            v1.push_back(make_pair(mnbal, bal));
        }
        else if(bal >= 0 && mnbal < 0){
            v2.push_back(make_pair(mnbal, bal));
        }
        else if(bal == mnbal){
            v4.push_back(make_pair(mnbal, bal));
        }
        else{
            v3.push_back(make_pair(mnbal, bal));
        }
    }
    int cnt = 0, k = 1;
    for(int i = 0; i < v1.size(); i++){
        cnt = cnt + v1[i].second;
    }
    sort(v2.rbegin(), v2.rend());
    for(int i = 0; i < v2.size(); i++){
        if(cnt + v2[i].first < 0){
            cout << "No";
            return 0;
        }
        cnt = cnt + v2[i].second;
    }
    sort(v3.rbegin(), v3.rend());
    for(int i = 0; i < v3.size(); i++){
        if(cnt + v3[i].first < 0){
            cout << "No";
            return 0;
        }
        cnt = cnt + v3[i].second;
    }
    for(int i = 0; i < v4.size(); i++){
        cnt = cnt + v4[i].second;
    }
    if(cnt == 0){
        cout << "Yes";
    }
    else{
        cout << "No";
    }
    return 0;
}