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 <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
using namespace std;
vector<pair<int, pair<int, int>>>first;
vector<pair<int, pair<int, int>>>second;
int N;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> N;
	int i;
	int su = 0;
	for (i = 0; i < N; i++)
	{
		string s;
		cin >> s;
		int j;
		int l = s.length();
		int cur = 0;
		int m = 0;
		for (j = 0; j < l; j++)
		{
			cur += s[j] == '(' ? 1 : -1;
			m = min(m, cur);
		}
		if (cur > 0)
		{
			first.push_back({ -m,{cur,l} });
		}
		else
		{
			second.push_back({ m - cur, {cur,l} });
		}
		su += cur;
	}
	if (su)
	{
		cout << "No";
		return 0;
	}
	sort(first.begin(), first.end());
	sort(second.begin(), second.end());
	for (i = 0; i < second.size(); i++)
	{
		first.push_back({ -second[i].first - second[i].second.first,second[i].second });
	}
	int nowp = 0;
	for (i = 0; i < N; i++)
	{
		if (nowp - first[i].first < 0)
		{
			cout << "No";
			return 0;
		}
		nowp += first[i].second.first;
	}
	cout << "Yes";
}