A bracket sequence is a string that is one of the following:

1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. 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`.

Examples

Input

2
)
(()


Output

Yes


Input

2
)(
()


Output

No


Input

4
((()))
((((((
))))))
()()()


Output

Yes


Input

3
(((
)
)


Output

No
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;

int N; string s[1001001];

int main()
{
	cin.tie(NULL); ios_base::sync_with_stdio(false);

	cin >> N;
	int ab = 0;
	vector<pair<int, int> > P, Q;
	for (int i = 0; i < N; i++){
		cin >> s[i];

		int b = 0, mn = 0;
		for (char c : s[i]){
			if (c == '(') b++;
			else b--;
			mn = min(mn, b);
		}

		ab += b;
		if (b >= 0) P.push_back({ -mn, -b });
		else Q.push_back({ b - mn, b });
	}

	if (ab){
		cout << "No" << endl;
		return 0;
	}

	sort(P.begin(), P.end());
	sort(Q.begin(), Q.end());

	for (auto &t : { P, Q }){
		int b = 0;
		for (auto &p : t){
			if (b - p.first < 0){
				cout << "No" << endl;
				return 0;
			}
			b -= p.second;
		}
	}

	cout << "Yes" << endl;
	return 0;
}
