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;

inline pair <int,int> read() {
	string s;
  cin >> s;
  int n = s.size(), miin = 0, sum = 0;
	for (int i = 0; i < n; i++) {
		sum += (s[i] == '(') ? 1 : -1;
		miin = min(miin, sum);
	}
	return {sum, miin};
}

inline bool comp(const pair<int,int> &a, const pair<int,int> &b) {
  return max(-a.second, -(a.first + b.second)) < max(-b.second,-(b.first + a.second));
}

int main () {
	int n; scanf ("%d",&n);
	
  vector <pair <int, int>> v;
  for (int i = 0; i < n; i++) v.push_back(read());

	sort(v.begin(),v.end(), comp);
	
  int sum = 0;

	for (int i = 0; i < n; i++) {
		if (sum + v[i].second < 0) return !printf ("No\n");
		sum += v[i].first;
	}

	if (sum == 0) printf ("Yes"); else printf ("No\n");
}