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 rep(i,n) for(int i=0;i<(n);++i)
#define Rep(i,n) for(int i=1;i<=(n);++i)
using LL = long long;
using ULL = unsigned long long;

char s[1000123];
pair<int, int> x[1000123], y[1000123];

int main()
{
	int n, c = 0, d = 0;
	cin >> n;
	rep(i, n) {
		int a = 0, b = 0;
		scanf("%s", s);
		for(int j = 0; s[j]; ++j) {
			if(s[j] == '(') {
				++b;
			}
			else {
				if(b) --b;
				else ++a;
			}
		}
		if(a <= b) x[c++] = {a, b};
		else y[d++] = {-b, a};
	}
	sort(x, x+c);
	sort(y, y+d);
	int cnt = 0;
	rep(i, c) {
		if(cnt >= x[i].first) cnt += x[i].second - x[i].first;
		else {
			puts("No");
			return 0;
		}
	}
	rep(i, d) {
		if(cnt >= y[i].second) cnt += -(y[i].second + y[i].first);
		else {
			puts("No");
			return 0;
		}
	}
	if(cnt == 0) puts("Yes");
	else puts("No");
	return 0;
}
