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;
int n,ct[1010101],nw[1010101],sum,al;
string s;
vector<pair<int,int> >v;
bool cmp(pair<int,int>a,pair<int,int>b)
{
	return min(a.second,a.first+b.second)>min(b.second,b.first+a.second);
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>s;
		int a=0;
		for(int j=0;j<s.size();j++)
		{
			if(s[j]=='(')
			{
				a++;
			}
			else
			{
				a--;
			}
			nw[i]=min(nw[i],a);
		}
		ct[i]=a;
		sum+=a;
		v.push_back(make_pair(ct[i],nw[i]));
	}
	if(sum!=0)
	{
		cout<<"No"<<endl;
		return 0;
	}
	sort(v.begin(),v.end(),cmp);
	int nn=0;
	for(int i=0;i<n;i++)
	{
		if(nn<-v[i].second)
		{
			cout<<"No"<<endl;
			return 0;
		}
		nn+=v[i].first;
	}
	cout<<"Yes"<<endl;
	return 0;
}
