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;

using ll = long long;

bool check (vector <pair<int, int>> a) {
	int cur = 0;
	for (auto i:a) {
		if (cur + i.first < 0) return 0;
		cur += i.second;
	}
	return 1;
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n, total = 0;
	cin >> n;
	vector <pair<int, int>> ls, rs;
	for (int i=0; i<n; i++) {
		string s;
		cin >> s;
		int mn = 0, cur = 0;
		for (char &i:s) {
			cur += (i == '(' ? 1 : -1);
			mn = min(mn, cur);
		}
		total += cur;
		if (cur > 0) {
			ls.push_back({mn, cur});
		} else {
			rs.push_back({mn-cur, -cur});
		}
	}
	sort(ls.rbegin(), ls.rend());
	sort(rs.rbegin(), rs.rend());
	if (!total && check(ls) && check(rs)) {
		cout << "Yes";
	} else {
		cout << "No";
	}
	return 0;
}
