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;
typedef long long ll;
bool solve() {
	int N;
	cin >> N;
	vector<string> S(N);
	for (auto& x : S) cin >> x;
	vector<pair<int,int>> a1;
	vector<pair<int,int>> a2;
	for (auto x : S) {
		int y = 0, z = 0;
		for (char c : x) {
			if (c == '(') {
				++ y;
			} else {
				-- y;
				z = min(z, y);
			}
		}
		if (y >= 0) {
			a1.push_back({-z,y});
		} else {
			int y2 = 0, z2 = 0;
			reverse(x.begin(), x.end());
			for (char c : x) {
				if (c == ')') {
					++ y2;
				} else {
					-- y2;
					z2 = min(z2, y2);
				}
			}
			a2.push_back({-z2,y2});
		}
	}
	sort(a1.begin(), a1.end());
	sort(a2.begin(), a2.end());
	int s1 = 0, s2 = 0;
	for (auto p : a1) {
		if (s1 < p.first) return false;
		s1 += p.second;
	}
	for (auto p : a2) {
		if (s2 < p.first) return false;
		s2 += p.second;
	}
	return s1 == s2;
}
int main() {
	cout << (solve() ? "Yes" : "No") << endl;
}
