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>
#define maxn 1000010
using namespace std;
struct node{
	int l,r;
}a[maxn];
int tot;
char s[maxn];
int cmp(node a,node b){
	return a.l-a.r>b.l-b.r;
}
int main(){
	int n,ln,rn;
	int i,j;
	ln=rn=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		int l=0,r=0;
		scanf("%s",s+1);
		for(j=1;s[j];j++)
			if(s[j]=='(')l++;
			else if(s[j]==')'){
				if(l)l--;
				else r++;
			}
		if(!l&&r)rn+=r;
		else if(l&&!r)ln+=l;
		else if(l&&r)a[++tot]=(node){l,r};	
	}
	sort(a+1,a+1+tot,cmp);
	for(int i=1;i<=tot;i++){
		if(ln<a[i].r)
			return printf("No\n")*0;
		ln-=a[i].r,ln+=a[i].l;
	}
	if(ln!=rn)return printf("No\n")*0;
	printf("Yes\n");
	return 0;
}