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>
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef pair <int, int> pii;

char buf[1000010];

bool cmp(pii a, pii b) {
	return max(-a.se, -a.fi - b.se) < max(-b.se, -b.fi - a.se);
}

int main () {
	int n; scanf("%d", &n);
	int tot = 0;
	vector <pii> p, q;
	for (int i = 0; i < n; i++) {
		scanf("%s", buf);
		int len = strlen(buf);
		int sum = 0, minj = 0;
		for (int i = 0; i < len; i++) {
			sum += buf[i] == '(' ? 1 : -1;
			minj = min(minj, sum);
		}
		tot += sum;
		if (sum >= 0) p.push_back(mp(minj, sum));
		else q.push_back(mp(sum, minj));
	}
	if (tot != 0) {
		printf("No\n");
		return 0;
	}
	sort(p.begin(), p.end());
	int sum = 0;
	while (p.size() && p.back().fi + sum >= 0) {
		sum += p.back().se;
		p.pop_back();
	}
	if (p.size()) {
		printf("No\n");
		return 0;
	}
	sort(q.begin(), q.end(), cmp);
	for (int i = 0; i < q.size(); i++) {
		if (sum + q[i].se < 0) {
			printf("No\n");
			return 0;
		}
		sum += q[i].fi;
	}
	printf("Yes\n");
	return 0;
}
