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;

const int M = 1e6 + 100;
int l[M], r[M];
int id[M], tp[M];

int main() {
	int n; cin >> n;
	for (int i = 0; i < n; i++) {
		string s; cin >> s;
		int x = 0, y = 0;
		for (char c : s) {
			if (c == '(') ++x;
			else if (x) --x;
			else ++y;
		}
		l[i] = x, r[i] = y, id[i] = i;
		if (y == 0) tp[i] = 1;
		else if (x == 0) tp[i] = 4;
		else if (x >= y) tp[i] = 2;
		else tp[i] = 3;
	}
	sort(id, id + n, [&] (int a, int b) {
		if (tp[a] != tp[b]) return tp[a] < tp[b];
		if (tp[a] == 2) {
			if (r[a] != r[b]) return r[a] < r[b];
			else return l[a] > l[b];
		}
		if (tp[a] == 3) return l[a] > l[b];
		return false;
	});
	int sum = 0;
	for (int i = 0; i < n; i++) {
		sum -= r[id[i]];
		if (sum < 0) break;
		sum += l[id[i]];
	}
	cout << (sum == 0 ? "Yes" : "No");
}