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,Z=0;
  cin>>N;
  vector<pair<int,int>> p;
  vector<pair<int,int>> q;
  int A=0;
  for(int i=0;i<N;i++){
    string S;
    cin>>S;
    int a=S.size(),b=0,c=0;
    for(int j=0;j<a;j++){
      if(S.at(j)=='('){
        b++;
      }
      else{
        b--;
      }
      if(c>b){
        c=b;
      }
    }
    if(b>=0&&c==0){
      A+=b;
    }
    else if(b>=0){
      p.push_back(make_pair(c,b));
    }
    else{
      q.push_back(make_pair(c-b,b*(-1)));
    }
  }
  sort(q.begin(), q.end());
  sort(p.begin(), p.end());
  reverse(p.begin(), p.end());
  int P=p.size(),Q=q.size();
  for(int i=0;i<P;i++){
    if(A+p.at(i).first>=0){
      A+=p.at(i).second;
    }
    else{
      Z=1;
      break;
    }
  }
  if(Z==0){
    for(int i=0;i<Q;i++){
      if(A-q.at(i).second+q.at(i).first>=0){
        A-=q.at(i).second;
      }
      else{
        Z=1;
        break;
      }
    }
  }
  if(Z==0&&A==0){
    cout<<"Yes"<<endl;
  }
  else{
    cout<<"No"<<endl;
  }
  
  
}