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 N=1e6+10;

struct rec {
	int x,y;
} a[N],b[N];
int la,lb;

bool cmp1(rec a,rec b) {return a.x<b.x;}
bool cmp2(rec a,rec b) {return a.y>b.y;}

int n,s0,s1,m;
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++;
		//-r +l
		if(r<l) a[++la]=(rec){r,l};
		else b[++lb]=(rec){r,l};
		s0+=r; s1+=l;
	}
	if(s0^s1) return puts("No"),0;
	sort(a+1,a+la+1,cmp1);
	sort(b+1,b+lb+1,cmp2);
	for(int i=1;i<=la;i++) {
		if(m<a[i].x) return puts("No"),0;
		m-=a[i].x-a[i].y;
	}
	for(int i=1;i<=lb;i++) {
		if(m<b[i].x) return puts("No"),0;
		m-=b[i].x-b[i].y;
	}
	puts("Yes"); return 0;
}
