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>
using namespace std;
const int maxn=1000005;
int n,a[maxn],b[maxn],id[maxn];char s[maxn];
bool cmp(int x,int y){
	if(b[x]>=0&&b[y]>=0) return a[x]<a[y];
	if(b[x]>=0&&b[y]<0) return 1;
	if(b[x]<0&&b[y]>=0) return 0;
	if(b[x]<0&&b[y]<0) return a[x]+b[x]>a[y]+b[y];
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%s",s+1);
		for (int j=1;s[j];j++)
		b[i]+=s[j]=='('?1:-1,
		a[i]=max(a[i],-b[i]);
		b[0]+=b[i];
	}
	if(b[0]){puts("No");return 0;}
	for (int i=1;i<=n;i++) id[i]=i;
	sort(id+1,id+1+n,cmp);
	for (int _i=1,i=id[_i],sum=0;_i<=n&&b[i]>=0;i=id[++_i]){
		if(sum<a[i]){puts("No");return 0;}
		sum+=b[i];
	}
	for (int _i=n,i=id[_i],sum=0;_i&&b[i]<0;i=id[--_i]){
		sum-=b[i];
		if(sum<a[i]){puts("No");return 0;}
	}
	puts("Yes");return 0;
}