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 <stdio.h>
#include <algorithm>
int a, b, n, g, x, y, f, fx, fy, now;
char s[1000001];
long long u[1000000];
long long d[1000000];
long long p;
long long q = 10000000;
int uu, dd;
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf(" %s", s);
		g = 0;
		x = 0;
		a = 0;
		while (s[g] != NULL)g++;
		for (int j = 0; j < g; j++) {
			if (s[j] == ')')x++;
			if (s[j] == '(')x--;
			if (x > a)a = x;
		}
		b = a - x;
		if (b < 0)b = 0;
		y += x;
		if (x <= 0) {
			p = a;
			p *= q;
			p -= x;
			u[uu] = p;
			uu++;
		}
		if (x > 0) {
			p = b;
			p *= q;
			p += x;
			d[dd] = p;
			dd++;
		}
	}
	if (y != 0) {
		printf("No\n");
		return 0;
	}
	std::sort(d, d + dd);
	std::sort(u, u + uu);
	for (int i = 0; i < uu; i++) {
		fx = u[i] % q;
		fy = u[i] / q;
		if (now < fy)f++;
		now += fx;
	}
	now = 0;
	for (int i = 0; i < dd; i++) {
		fx = d[i] % q;
		fy = d[i] / q;
		if (now < fy)f++;
		now += fx;
	}
	if (f)printf("No\n");
	else printf("Yes\n");
}