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)

#pragma GCC optimize("Ofast", "unroll-loops")

#include <bits/stdc++.h>

using namespace std;

pair<int, int> calc_mn_total(string& s){
    int mn = 0, total = 0;
    int sz = s.length();
    for (int i = 0; i < sz; ++i){
        if (s[i] == '(')
            ++total;
        else{
            --total;
            mn = min(mn, total);
        }
    }
    return pair<int, int>(mn, total);
}

int main(){
    int N; cin >> N;
    vector<pair<int, int>> s(N);
    for (int i = 0; i < N; ++i){
        string si; cin >> si;
        s[i] = calc_mn_total(si);
    }

    vector<pair<int, int>> zero_pos, neg;
    for (auto& p : s){
        if (p.first == 0) zero_pos.push_back(p);
        else neg.push_back(p);
    }

    int now = 0;
    for (auto p : zero_pos)
        now += p.second;

    sort(
        neg.begin(), neg.end(),
        [&](auto p1, auto p2){
            return (p1.second - p1.first) > (p2.second - p2.first);
        }
    );
    for (auto p : neg){
        if (now + p.first < 0){
            cout << "No" << endl;
            return 0;
        }
        now += p.second;
    }

    string res = (now == 0) ? "Yes" : "No";
    cout << res << endl;
    
    return 0;
}