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;


int main(){
  int N;
  cin >> N;
  vector<vector<int>> H(N, vector<int>(2, 0));
  for(int ii=0; ii<N; ii++){
    string s;
    cin >> s;
    for(auto c : s){
      if(c=='(') H[ii][0]++;
      else H[ii][0]--;
      H[ii][1] = min(H[ii][1], H[ii][0]);
    }
  }

  sort(H.begin(), H.end(), [](vector<int>& a, vector<int>& b){
    if(a[0]>0 && b[0]>0) return a[1] > b[1];
    if(a[0]<=0 && b[0]<=0) return a[1]-a[0] < b[1]-b[0];
    if(a[0]>0 && b[0]<=0) return true;
    if(a[0]<=0 && b[0]>0) return false;
  });

  long long int height=0;
  for(auto& h : H){
    if(height + h[0] < 0){
      cout << "No" << endl;
      return 0;
    }
    if(height + h[1] < 0){
      cout << "No" << endl;
      return 0;
    }
    height += h[0];
  }
  cout << (height==0?"Yes":"No") << endl;
  return 0;
}
