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 <bits/stdc++.h>
using namespace std;

int main(){
	int N, i, h_sum = 0;
	scanf("%d", &N);
	vector<int> h(N, 0), d(N, 0);
	for(i = 0; i < N; i++){
		string sub;
		cin >> sub;
		for(int j = 0; j < sub.size(); j++){
			if(sub[j] == '('){
				h[i]++;
			}
			else{
				h[i]--;
			}
			d[i] = max(d[i], -h[i]);
		}
		h_sum += h[i];
	}
	if(h_sum != 0){
		printf("No\n");
		return 0;
	}
	vector<pair<int, int>> dh_plus(0), dh_minus(0);
	for(i = 0; i < N; i++){
		if(h[i] >= 0){
			dh_plus.push_back(make_pair(d[i], h[i]));
		}
		else{
			dh_minus.push_back(make_pair(d[i] + h[i], -h[i]));
		}
	}
	sort(dh_plus.begin(), dh_plus.end());
	sort(dh_minus.begin(), dh_minus.end());
	int H = 0;
	for(i = 0; i < dh_plus.size(); i++){
		if(dh_plus[i].first > H){
			printf("No\n");
			return 0;
		}
		H += dh_plus[i].second;
	}
	H = 0;
	for(i = 0; i < dh_minus.size(); i++){
		if(dh_minus[i].first > H){
			printf("No\n");
			return 0;
		}
		H += dh_minus[i].second;
	}
	printf("Yes\n");
	return 0;
}