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 <cstdio>
#include <algorithm>
#include <utility>
using namespace std;

const int N = 1000000;
int n;
pair<int, int> a1[N];
pair<int, int> a2[N];
int n1, n2;
char s[N + 1];

bool test(pair<int, int> *a, int n) 
{
	sort(a, a + n);
	int p = 0;
	for (int i = 0; i < n; ++i) {
		if (a[i].first > p) return false;
		p += a[i].second;
	}
	return true;
}

int main() 
{
	scanf("%d", &n);
	int b = 0;
	for (int i = 0; i < n; ++i) {
		scanf("%s", s);
		int f = 0;
		int l = 0;
		for (int j = 0; s[j]; ++j) {
			f += (s[j] == '(') * 2 - 1;
			l = min(l, f);
		}
		b += f;
		if (f >= 0) {
			a1[n1] = make_pair(-l, f);
			++n1;
		}
		else {
			a2[n2] = make_pair(-l + f, -f);
			++n2;
		}
	}
	if (!b && test(a1, n1) && test(a2, n2)) printf("Yes\n");
	else printf("No\n");
	return 0;
}
