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;
typedef int64_t ll;


int main() {
  int n;
  cin >> n;
  vector<int> plus(n,0),low(n,0);
    vector<tuple<bool,int,int>> info;
  for(int i=0;i<n;i++){
    string s;
    cin >> s;
    for(int j=0;j<(int)s.size();j++){
      if(s[j]=='(') plus[i]++;
      else plus[i]--;
      low[i]=min(low[i],plus[i]);
    }
    info.push_back({plus[i]<0,(plus[i]<0?low[i]-plus[i]:-low[i]),plus[i]});
  }
  sort(info.begin(),info.end());
  int h=0;
  for(auto [a,b,c] : info){
    if(a && h+b+c<0 || !a && h-b<0){cout << "No" << endl;return 0;}
    else h+=c;
  }
  if(h!=0){cout << "No" << endl;return 0;}
  cout << "Yes" << endl;
}
