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;
#define int long long
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);cout.tie(0);
  int no_of_input;
  string input;
  vector<pair<int,int> > vect1;
  vector<pair<int,int> > vect2;
  int curr = 0;
  int neg = 0;
  cin >> no_of_input;
  for(int i=0;i<no_of_input;i++){
    cin >> input;
    int currmini = INT_MAX,currsum = 0;
    for(int j=0;j<input.length();j++){
      if(input.at(j)=='('){
        currsum++;
      }
      else{
        currsum--;
      }
      currmini = min(currmini,currsum);
    }
    if(currmini>=0){
      curr += currsum;
    }
    else if(currmini==currsum){
      neg += currsum;
    }
    else{
      vect2.push_back({(currsum),currmini});
    }
  }
  sort(vect2.begin(),vect2.end());
  reverse(vect2.begin(),vect2.end());
  for(int i=0;i<vect2.size();i++){
    if(abs(vect2[i].second)>curr){
      cout << "No";
      return 0;
    }
    curr += vect2[i].first;
  }
  curr += neg;
  if(curr==0){
    cout << "Yes";
  }
  else{
    cout << "No";
  }
  
}