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;

typedef pair<int,int> PII;
const int M=1e6+5;
char s[M];

struct Node
{
	int r,sum;
	Node(int r=0,int sum=0)
		:r(r),sum(sum){}
	bool operator<(const Node &B) const
	{
		return r<B.r;
	}
};
vector<Node> vec1,vec2;

void check()
{
	int m=strlen(s);
	int r=0,sum=0;
	for(int i=0;i<m;i++)
	{
		if(s[i]=='(') sum++;
		else
		{
			if(sum) sum--;
			else r++;
		}
	}
	sum-=r;
	if(sum>=0)
	{
		vec1.push_back(Node(r,sum));
		return;
	}
	r=0;sum=0;
	for(int i=m-1;i>=0;i--)
	{
		if(s[i]==')') sum++;
		else
		{
			if(sum) sum--;
			else r++;
		}
	}
	sum-=r;
	vec2.push_back(Node(r,sum));
}

int main()
{
	int n; scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%s",s);
		check();
	}
	sort(vec1.begin(),vec1.end());
	sort(vec2.begin(),vec2.end());
	int suml=0,sumr=0,f=1;
	for(auto p:vec1)
	{
		if(suml<p.r)
		{
			f=0;
			break;
		}
		suml+=p.sum;
	}
	for(auto p:vec2)
	{
		if(sumr<p.r)
		{
			f=0;
			break;
		}
		sumr+=p.sum;
	}
	if(suml!=sumr) f=0;
	printf("%s\n",f?"Yes":"No");
	return 0;
}