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;
const int INF=0x3f3f3f3f; 
typedef long long ll;

 char a[1000005]; 
 
int main(){
	int n,i,j,len,L=0,R=0;
	int sum1=0,sum2=0;
	int left=0,right=0;
	int minl=INF,minr=INF;
	scanf("%d",&n);
	while(n--){
		scanf("%s",&a);
		int l=0,r=0,sum=0;
		len=strlen(a);
		for(i=0;i<len;i++){
			if(a[i]=='('){
				sum++;
				sum1++;
			}
			else {
				sum2++;
				if(sum==0)l++;
				else sum--;
			}
		}
		sum=0;
		for(i=len-1;i>=0;i--){
			if(a[i]==')')sum++;
			else {
				if(sum==0)r++;
				else sum--;
			}
		}
		if(l==0&&r==0)continue;
		else if(l==0&&r!=0)R+=r;
		else if(l!=0&&r==0)L+=l;
		
			
			else {
				minl=min(minl,l);
				minr=min(minr,r);
				left+=l;
				right+=r;
			}
		
	}
	if((minl>R||minr>L)&&(minl!=INF)&&(minr!=INF))cout<<"No"<<endl;
	else if(left+L!=right+R)cout<<"No"<<endl;
	else {
		if(L&&R)cout<<"Yes"<<endl;
		else if(L==0&&R==0&&left==0&&right==0)cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}

    return 0;
}
