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;
using ll = long long;

const int N = 1e6 + 5, MOD = 1e9 + 7;
const ll INF = 1e18 + 17;

int n, bal[N];

string s[N];

void solve() {
	cin >> n;
	vector <pair <int, int> > v;
	string res;
	for (int i = 1; i <= n; ++i) {
		cin >> s[i];
		int sz = s[i].size();
		int bal = 0, mn = sz;
		for (int j = 0; j < sz; ++j) {
			if (s[i][j] == '(') ++bal;
			else bal--;
			mn = min(mn, bal);
		}
		v.push_back({mn, -i});
	}
	sort(v.begin(), v.end());
	reverse(v.begin(), v.end());
	for (int i = 0; i < n; ++i) {
		res += s[-v[i].second];
	}
	//cout << res << '\n';
	int bal = 0;
	for (int i = 0; i < res.size(); ++i) {
		if (res[i] == '(') ++bal;
		else bal--;
		if (bal < 0) {
			cout << "No";
			return;
		}
	}
	if (!bal) cout << "Yes";
	else cout << "No";
}

main() {
	ios :: sync_with_stdio(0);		
	cin.tie(0);
	int tt = 1;
	//cin >> tt;
	while (tt--) {
		solve();
		cout << '\n';
	} 
	return 0;
}