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;
typedef long long ll;
typedef unsigned long long ull;

struct num{
    int a;
    int b;
    bool operator<(const num& right) const{
        return a < right.a;
    }
};

int main(void)
{
    int N;
    cin >> N;
    vector<string> S(N);
    for(int i=0;i<N;i++) {
        cin >> S[i];
    }
    // A, B
    vector<num> numV(N);
    for(int i=0;i<N;i++){
        string s = S[i];
        int a=0, b=0;
        for(unsigned int j=0;j<s.size();j++){
            if(s[j] == '(')b++;
            else {
                if(b==0) a++;
                else b--;
            }
        }
        num n = {a, b};
        numV[i] = n;
    }
    sort(numV.begin(), numV.end());
    //  check
    ll total_b = 0;
    for(int i=0;i<N;i++){
        total_b -= numV[i].a;
        if (total_b<0){
            cout << "No" << endl;
            return 0;
        }
        total_b += numV[i].b;
    }
    if(total_b==0) {
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
    return 0;
}