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;
#define int long long
const int M = 1e6 + 5;
int n;

bool check(string ans)
{
	int cur = 0;
	for(auto &c : ans)
	{
		cur += (c == '(' ? 1 : -1);
		if(cur < 0)
			return 0;
	}
	return cur == 0;
}
 
int32_t main() 
{
	ios_base::sync_with_stdio(0);	cin.tie(0);	cout.tie(0);
	cin >> n;
	vector <string> s(n);
	vector <pair <int, int>> pos, neg;
	string ans = "";
	for(int i = 0; i < n; i++)
	{
		cin >> s[i];
		int bal = 0, mn = 1e9;
		for(auto &c : s[i])
		{
			bal += (c == '(' ? 1 : -1);
			mn = min(mn, bal);
		}
		if(mn < 0)
		{
			if(bal > 0)	pos.push_back({mn, i});
			else		neg.push_back({mn - bal, i});
		}
		else	ans += s[i];
	}
	sort(pos.rbegin(), pos.rend());
	sort(neg.begin(), neg.end());
	for(auto &i : pos)
		ans += s[i.second];
	for(auto &i : neg)
		ans += s[i.second];
	// cout << ans << "\n";
	cout << (check(ans) ? "Yes" : "No"); 
    	return 0;
}