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 <string>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
	int N;
	cin >> N;

	vector<pair<int, int>> former, latter;

	for (int i = 0; i < N; i++){
		string S;
		cin >> S;

		int count = 0;
		int A = 1e9;
		for (int j = 0; j < S.size(); j++){
			if (S[j] == '('){
				count++;
			} else {
				count--;
			}
			A = min(A, count);
		}
		if(count >= 0){
			former.push_back(make_pair(-A, count));
		} else {
			latter.push_back(make_pair(count - A, -count));
		}
	}
	sort(former.begin(), former.end());
	sort(latter.begin(), latter.end());

	int count1 = 0;
	for (int i = 0; i < former.size(); i++){
		if(count1 < former[i].first){
			cout << "No" << endl;
			return 0;
		}
		count1 += former[i].second;
	}

	int count2 = 0;
	for (int i = 0; i < latter.size(); i++){
		if(count2 < latter[i].first){
			cout << "No" << endl;
			return 0;
		}
		count2 += latter[i].second;
	}

	if (count1 == count2){
		cout << "Yes" << endl;
	} else {
		cout << "No" << endl;
	}
	return 0;
}
