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)

/*
 *  Language:   C++
 *  Author:     Debajyoti Dasgupta
 */

#include <bits/stdc++.h>

using namespace std;
#define int long long

signed main() {
    int n;
    cin >> n;
    vector<string> vs(n);
    for (int i = 0; i < n; ++i) {
        cin >> vs[i];
    }
    vector<vector<int>> tmp(n);
    int cnt;
    for (int i = 0; i < n; ++i) {
        cnt = 0;
        for (auto j  : vs[i]) {
            if (j == '(')cnt++;
            else cnt--;
            tmp[i].emplace_back(cnt);
        }
    }
    vector<tuple<int, int, int>> data(n);
    for (int i = 0; i < n; ++i) {
        get<0>(data[i]) = *min_element(tmp[i].begin(), tmp[i].end()) * (-1);
        get<1>(data[i]) = *max_element(tmp[i].begin(), tmp[i].end());
        get<2>(data[i]) = i;
    }
    sort(data.begin(), data.end());
    string s;
    for (int i = 0; i < n; ++i) {
        s += vs[get<2>(data[i])];
    }
    stack<char> st;
    for (auto i :s) {
        if (i == '(')st.push('(');
        else {
            if (st.empty()) {
                cout << "No" << endl;
                return 0;
            } else st.pop();
        }
    }
    if (st.empty()) cout << "Yes" << endl;
    else cout << "No" << endl;
}