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 <list>
#include <queue>
#include <algorithm>
#define rep(i, n) for(i = 0; i < (n); i++)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define MOD 1000000007
#define PI 3.14159265358979323846
#define INF 1 << 30

using namespace std;
typedef long long ll;
typedef pair<int, int> pp;

int Judge(vector<pp> & d) {
	int i, n = 0;
	rep(i, d.size()) {
		if (n + d[i].first < 0)
			return -1;
		n += d[i].second;
	}
	return n;
}

int main(void) {
	int num, i, j;
	string s;
	cin >> num;
	vector<pp> a, b;
	rep(i, num) {
		pp n = { 0,0 };
		cin >> s;
		rep(j, s.size()) {
			if (s[j] == '(')
				n.second++;
			else
				n.second--;
			chmin(n.first, n.second);
		}
		if (n.second >= 0)
			a.push_back(n);
		else
			b.push_back({ n.first - n.second, -n.second });
	}
	sort(a.begin(), a.end(), greater<pp>());
	sort(b.begin(), b.end(), greater<pp>());
	i = Judge(a);
	j = Judge(b);
	if (i == j && i >= 0)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}