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;

#define int long long

const int N = (int)1e6 + 5, MOD = (int)1e9 + 7, INF = (int)3e18;

string s[N];

main() {
    ios :: sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    string ans;
    for (int i = 1; i <= n; ++i) {
        cin >> s[i];
    }
    int sum = 0;
    vector <pair <int, string> > v;
    string z;bool was = 0;
    for (int i = 1; i <= n; ++i) {
        int bal = 0;
        int mn = INF;
        for (auto c : s[i]) {
            if (c == '(') {
                ++bal;
            }
            else {
                bal--;
            }
            mn = min(mn, bal);
        }
        sum += bal;
        if (bal < 0 && s[i].back() == ')' && !was)  {
        z += s[i];
            was = 1;
        }
        else
            v.push_back(make_pair(mn, s[i]));
    }
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    string t;
    for (auto x : v) t += x.second;
    t += z;
    int bal = 0;
    for (auto x : t) {
        if (x == '(') ++bal;
        else bal--;
        if (bal < 0) {
            cout << "No";
            return 0;
        }   
    }
    if (bal == 0) cout << "Yes";
    else {
        cout << "No";
    }
    return 0;
}