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 <string>
#include <algorithm>

using namespace std;

bool ok(const vector<pair<int, int>>& v)
{
	int p = 0;
	for (auto& x : v) {
		if (p + x.first < 0) {
			return false;
		}
		p += x.second;
	}
	return true;
}

int main()
{
	int N;
	cin >> N;
	vector<string> S(N);
	for (auto& x : S) {
		cin >> x;
	}

	vector<pair<int, int>> L, R;
	int total = 0;
	for (const auto& x : S) {
		int h = 0;
		int b = 0;
		for (auto c : x) {
			if (c == '(') {
				++h;
			}
			else {
				--h;
			}
			b = min(b, h);
		}

		if (h > 0) {
			L.emplace_back(b, h);
		}
		else {
			R.emplace_back(b - h, -h);
		}

		total += h;
	}

	sort(L.rbegin(), L.rend());
	sort(R.rbegin(), R.rend());
	if (total == 0 && ok(L) && ok(R)) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
}