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 nm = 1e6;
int n, l, r, cnt, ls, rs;
;
vector<pair<int, int>> b, c;

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		string s;
		cin >> s;
		cnt = 0, l = 0, r = 0;
		for (int i = 0; i < (int) s.size(); i++) {
			if (s[i] == '(')
				cnt--;
			else
				cnt++;
			l = max(cnt, l);
		}
		cnt = 0;
		for (int i = s.size() - 1; i >= 0; i--) {
			if (s[i] == ')')
				cnt--;
			else
				cnt++;
			r = max(cnt, r);
		}
		if (l == 0) {
			ls += r;
		} else if (r == 0)
			rs += l;
		else if (r - l >= 0)
			b.emplace_back(l, r);
		else
			c.emplace_back(r, l);
	}
	sort(b.begin(), b.end());
	sort(c.begin(), c.end());
	for (auto &x : b) {
		if (x.first <= ls)
			ls += x.second - x.first;
		else {
			cout << "No" << endl;
			return 0;
		}
	}
	for (auto &x : c) {
		if (x.first <= rs)
			rs += x.second - x.first;
		else {
			cout << "No" << endl;
			return 0;
		}
	}
	cout << (ls == rs ? "Yes" : "No") << endl;
}
