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<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+10;

int n,left,right,s0,s1;
struct node {
	int x,y;
	bool operator <(node b) const {
		return x - y>b.x - b.y; //更大在前面 
	}
} a[N];

char s[N];

int main() {
	scanf("%d",&n);
	for(int i=1,l,r;i<=n;i++) {
		scanf("%s",s+1); l=r=0;
		for(int j=1;s[j];j++)
			if(s[j]=='(') l++;
			else {
				if(l) l--;
				else r++;
			}
		if(!l&&r) right+=r;
		else if(!r&&l) left+=l;
		else a[i]=(node){l,r};
		s0+=l;s1+=r;
	}
	if(s0^s1) puts("No"),exit(0);
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++) {
		if(left<a[i].y) puts("No"),exit(0);
		left-=a[i].y-a[i].x;
	}
	if(left^right)puts("No");
	else puts("Yes"); return 0;
}
