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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <fstream>
#include <sstream>
#include <iomanip>
#define ll long long

using namespace std;

long long MOD = 1000000007;

int main(){
    int N;
    cin >> N;
    int tmp=0;
    vector<vector<int>> vec1(0);
    vector<vector<int>> vec2(0);
    for(int i=0; i<N; i++){
        string S;
        cin >> S;
        int minl,l;
        minl=0;l=0;
        for(int j=0; j<(int)S.size(); j++){
            if(S[j]=='('){
                l++;
            }else{
                l--;
                minl=min(minl,l);
            }
        }
        if(minl>=0){
            tmp+=l;
        }else if(l>0){
            vec2.push_back({minl,l});
        }else{
            vec1.push_back({l-minl,l});
        }
    }

    sort(vec2.rbegin(),vec2.rend());
    sort(vec1.rbegin(),vec1.rend());
    auto it=vec2.begin();
    for(int i=0; i<(int)vec1.size(); i++){
        while(it!=vec2.end() && tmp+(*it)[0]>=0){
            tmp+=(*it)[1];
            it++;
        }
        if(tmp-vec1[i][0]+vec1[i][1]<0){
            cout << "No" << endl;
            return 0;
        }else{
            tmp+=vec1[i][1];
        }
    }
    if(tmp==0){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
}