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;

#define ll long long
#define mp make_pair
#define pb push_back

#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, n) for (int i = 0; i < (int)(n); ++ i)

const int mxn = 1e6 + 5;
int n;
char s[mxn];
int bg[mxn], ln[mxn];
int a[mxn], b[mxn];
int id[mxn];

bool comp(int x, int y) {
	return min(-a[x], -a[x] + b[x] - a[y]) > min(-a[y], -a[y] + b[y] - a[x]);
}

void NO() {
	puts("No");
	exit(0);
}

int main() {
	scanf("%d", &n);
	rep(i, n) {
		scanf("%s", s + bg[i]);
		ln[i] = strlen(s + bg[i]);
		rep(j, ln[i]) {
			if (s[bg[i] + j] == '(') {
				++ b[i];
			} else {
				if (b[i]) -- b[i];
				else ++ a[i];
			}
		}
		bg[i + 1] = bg[i] + ln[i];
		id[i] = i;
	}
	sort(id, id + n, comp);
	int A = 0;
	rep(xx, n) {
		int i = id[xx];
		A -= a[i];
		if (A < 0) NO();
		A += b[i];
	}
	if (A != 0) NO();
	puts("Yes");
	return 0;
}
