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>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
vector<P> d,u;

int main()
{
	int N;
	cin >> N;
	rep(i,N)
	{
		string S;
		cin >> S;
		int t = 0, temp = 0;
		rep(j,S.size())
		{
			if (S[j] == '(') temp--;
			else temp++;
			t = max(t,temp);
			//b = min(b,temp);
		}
		if (temp <= 0)
		{
			d.push_back(P(t,temp));
		}
		else
		{
			u.push_back(P(t-temp,temp));
		}
	}
	sort(d.begin(), d.end());
	sort(u.begin(), u.end());
	reverse(u.begin(), u.end());
	int now = 0;
	rep(i,d.size())
	{
		now += d[i].first;
		if (now > 0)
		{
			cout << "No" << endl;
			return 0;
		}
		now = now - d[i].first + d[i].second;
	}
	rep(i,u.size())
	{
		now += u[i].first + u[i].second;
		if (now > 0)
		{
			cout << "No" << endl;
			return 0;
		}
		now -= u[i].first;
	}
	if (now == 0) cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}