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;
#define ll long long
#define all(V) V.begin(),V.end()
#define MOD 998244353

int main() {
	int N, s;
	cin >> N;
	string S;
	vector<pair<int, int>> VM, VP;
	int le, res;
	for (int i = 0;i < N;i++) {
		cin >> S;
		s = S.size();
		le = 0, res = 0;
		for (int j = 0;j < s;j++) {
			if (S[j] == '(') {
				res++;
			}
			else {
				res--;
			}
			le = min(le, res);
		}
		if (res < 0) {
			VM.push_back({ res - le,-res });
		}
		else {
			VP.push_back({ -le,res });
		}
	}
	sort(all(VM));
	sort(all(VP));
	res = 0;
	for (auto x : VP) {
		if (res - x.first < 0) {
			cout << "No" << endl;
			return 0;
		}
		res += x.second;
	}
	le = 0;
	for (auto x : VM) {
		if (le - x.first < 0) {
			cout << "No" << endl;
			return 0;
		}
		le += x.second;
	}
	if (res != le) {
		cout << "No" << endl;
	}
	else {
		cout << "Yes" << endl;
	}
}
