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;
  int p=0;
  int m=0;
  vector<vector<int>> vp;
  vector<vector<int>> vm;
  for(int _=0;_<n;_++){
    string s;
    cin>>s;
    int x=0;
    int v=0;
    for(int i=0;i<(int)s.size();i++){
      x+=(s.at(i)=='('?1:-1);
      v=min(v,x);
    }
    if(x>0){
      if(v==0)
        p+=x;
      else
        vp.push_back({v,x});
    }else{
      if(v==x)
        m+=x;
      else
        vm.push_back({v,x});
    }
  }
  sort(vp.begin(),vp.end(),greater<vector<int>>());
  sort(vm.begin(),vm.end());
  int x=p;
  for(int i=0;i<(int)vp.size();i++){
    if(x+vp.at(i).at(0)<0){
      cout<<"No"<<endl;
      return 0;
    }
    x+=vp.at(i).at(1);
  }
  for(int i=0;i<(int)vm.size();i++){
    if(x+vm.at(i).at(0)<0){
      cout<<"No"<<endl;
      return 0;
    }
    x+=vm.at(i).at(1);
  }
  cout<<(x+m==0?"Yes":"No")<<endl;
}