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>
//#include "stdc++.h"
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = s; i < (int)(n); i++)
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = max(a, b)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

using namespace std;
using ll = long long;
using pii = pair<int, int>;

bool	check(vector<pii> s)
{
	int h = 0;
	for (pii p : s) {
		int b = h + p.first;
		if (b < 0) return false;
		h += p.second;
	}
	return true;
}

void	solve()
{
	int n;
	cin >> n;
	vector<pii> ls, rs;
	int total = 0;
	while (n--) {
		string s;
		cin >> s;
		int h = 0, b = 0;;
		for (char c : s) {
			if (c == '(') 
				h++;
			else
				h--;
			b = min(b, h);
		}
		if (h > 0) ls.emplace_back(b, h);
		else rs.emplace_back(b - h, -h);
		total += h;
	}
	sort(rall(ls));
	sort(rall(rs));
	if (check(ls) && check(rs) && !total)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
}

int		main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	solve();
}