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;
vector<pair<int,int> > v[2];
int main()
{
	int n,sum=0;
	cin >> n;
	while (n--)
	{
		string s;
		cin >> s;
		int bal=0,mn=0;
		for (char c:s)
		{
			bal+=(c=='(');
			bal-=(c==')');
		}
		sum+=bal;
		if (bal>=0)
		{
			bal=0;
			for (char c:s)
			{
				bal+=(c=='(');
				bal-=(c==')');
				mn=min(mn,bal);
			}
			v[0].push_back({mn,bal});
		}
		else
		{
			reverse(s.begin(),s.end());
			bal=0;
			for (char c:s)
			{
				bal-=(c=='(');
				bal+=(c==')');
				mn=min(mn,bal);
			}
			v[1].push_back({mn,bal});
		}
	}
	if (sum)
	{
		cout << "No";
		return 0;
	}
	for (int i=0;i<2;i++)
	{
		sort(v[i].begin(),v[i].end(),greater<pair<int,int> >());
		int cur=0;
		for (auto p:v[i])
		{
			if (cur+p.first<0)
			{
				cout << "No";
				return 0;
			}
			cur+=p.second;
		}
	}
	cout << "Yes";
}