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;

vector <pair<int, int> > a[2];
char s[1000005];

bool cmp(pair<int, int> u, pair<int, int> v) {
	return u.second - u.first < v.second - v.first;
}

int main() {
	int n;
	scanf("%d", &n);
	
	for (int i = 0; i < n; i++) {
		scanf("%s", s);
		
		int cur = 0;
		int cost = 0;
		int l = strlen(s);
		
		for (int j = 0; j < l; j++) {
			if (s[j] == '(') cur++;
			else cur--;
			cost = max(cost, -cur);
		}
		
		if (cur >= 0) a[0].push_back(make_pair(cost, cur));
		else a[1].push_back(make_pair(cost, -cur));
	}
	
	sort(a[0].begin(), a[0].end());
	sort(a[1].begin(), a[1].end(), cmp);
	
	int now = 0;
	bool gg = 0;
	
	for (int i = 0; i < a[0].size(); i++) {
		if (now < a[0][i].first) gg = 1;
		now += a[0][i].second;
	}
	
	for (int i = 0; i < a[1].size(); i++) {
		if (now < a[1][i].first) gg = 1;
		now -= a[1][i].second;
	}
	
	if (gg || now != 0) printf("No\n");
	else printf("Yes\n");
}
