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>
#define Int int64_t

using namespace std;

int main() {
	int N;
	cin >> N;
	vector<string> s(N);
	for (int i = 0; i < N; ++i) { cin >> s[i]; }

	vector<pair<int, int>> a, b;
	for (int i = 0; i < N; ++i) {
		int sum = 0, min_s = 1e8;
		for (auto c : s[i]) {
			if (c == '(') { ++sum; }
			else { --sum; }
			min_s = min(min_s, sum);
		}
		if (sum >= 0) { a.emplace_back(-min_s, i); }
		else { b.emplace_back(min_s - sum, i); }
	}

	sort(a.begin(), a.end());
	sort(b.begin(), b.end());
	int sum = 0;
	for (auto p : a) {
		for (auto c : s[p.second]) {
			if (c == '(') { ++sum; }
			else { --sum; }
			if (sum < 0) {
				cout << "No\n";
				return 0;
			}
		}
	}
	for (auto p : b) {
		for (auto c : s[p.second]) {
			if (c == '(') { ++sum; }
			else { --sum; }
			if (sum < 0) {
				cout << "No\n";
				return 0;
			}
		}
	}
	cout << (sum == 0 ? "Yes" : "No") << endl;
	
	return 0;
}
