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>
using namespace std;

bool compare_second(pair<int, int> a, pair<int, int> b) {
	if (a.second == b.second) return a.first > b.first;
	return a.second > b.second;
}

int main() {
	int n;
	cin >> n;
	vector<pair<int, int>> A, B;
	for (int i = 0; i < n; i++) {
		string s;
		cin >> s;
		int m = 0;
		int sum = 0;
		for (int j = 0; j < s.size(); j++) {
			if (s[j] == ')') {
				sum++;
				m = max(m, sum);
			} else {
				sum--;
			}
		}
		int x = m, y = m - sum;
		if (x == 0 && y == 0) continue;
		if (y - x >= 0) A.emplace_back(x, y);
		else B.emplace_back(x, y);
	}
	sort(A.begin(), A.end());
	sort(B.begin(), B.end(), compare_second);
	A.insert(A.end(), B.begin(), B.end());
	int s = 0;
	for (auto p : A) {
		int x = p.first, y = p.second;
		if (s - x >= 0) {
			s = s - x + y;
		} else {
			cout << "No" << endl;
			return 0;
		}
	}
	if (s == 0) cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}