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,in=0,out=0,cant=0;
    cin >> N;
    for(int i=0;i<N;i++){
        string S;
        cin >> S;
        int inn=0,outt=0;
        for(int j=0;j<S.size();j++){
            if(S[j]=='('){
                inn++;
                if(inn<=outt){
                    inn++;
                    outt++;
                }
            }else{
                if(inn==0){
                    outt++;
                }else{
                    inn--;
                }
            }
        }

        if(inn!=0 || outt != 0){
            cant++;
        }

        in += inn;
        out += outt;
    }

    if(in == out && 1!=cant%2){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
}