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<algorithm>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<cstring>
#include<queue>
#include<stack>
#include<numeric>
using namespace std;
#define INF (1 << 30) - 1
#define LINF 1LL << 60
const int MOD = 1000000007;
using ll = long long;
using P = pair<int, int>;

bool is_not_minus(vector<P> a){
	int h = 0;
	for(P p : a){
		if(h + p.first < 0)return false;
		h += p.second;
	}
	return true;
}

int main(){
  int n;
  cin >> n;
  vector<P> plus, minus;
  int total = 0;
  for(int i = 0; i < n; ++i){
  	string s;
  	cin >> s;
  	int h = 0, bottom = 0;
  	for(char c : s){
  		if(c == '(')++h;
  		else --h;
  		bottom = min(bottom, h);
  	}
  	if(h > 0)plus.emplace_back(bottom, h);
  	else minus.emplace_back(bottom - h, -h);
  	total += h;
  }

  sort(plus.rbegin(), plus.rend());
  sort(minus.rbegin(), minus.rend());

  if(total == 0 && is_not_minus(plus) && is_not_minus(minus)){
  	cout << "Yes" << endl;
  }
  else cout << "No" << endl;
  return 0;
}
