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,tot;
struct Str
{
	int a,b,c;
	inline bool operator < (const Str& b)const
	{
		if(1ll*c*b.c<=0) return c>b.c;
		if(c<0) return this->b>b.b;
		return a<b.a;
	}
}A[maxn];
inline Str read()
{
	Str ret=(Str){0,0};char ch=getchar();
	while(ch!='('&&ch!=')') ch=getchar();
	while(ch=='('||ch==')')
	{
		if(ch==')') ret.b?ret.b--:ret.a++;
		else ret.b++;
		ch=getchar();
	}
	ret.c=ret.b-ret.a;
	return ret;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		A[i]=read();
		if(A[i].a==0)
		{
			tot+=A[i].b;
			i--;n--;continue;
		}
	}
	sort(A+1,A+1+n);
	for(int i=1;i<=n;i++)
	{
		tot-=A[i].a;
		if(tot<0)
		{
			printf("No\n");
			return 0;
		}
		tot+=A[i].b;
	}
	printf("%s\n",tot?"No":"Yes");
	return 0;
}