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 long long ll;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=998244353;

int main(){
  int n;
  cin>>n;
  string s[n];
  rep(i,0,n){
    cin>>s[i];
  }
  vector<P> left;
  vector<P> right;
  rep(i,0,n){
    int worst=0;
    int las=0;
    rep(j,0,s[i].size()){
      if(s[i][j]=='(') las++;
      else las--;
      worst=min(worst,las);
    }
    if(las>0){
      left.push_back(P(worst,las));
    }
    else{
      right.push_back(P(worst-las,-las));
    }
  }
  sort(left.begin(),left.end());
  reverse(left.begin(),left.end());
  sort(right.begin(),right.end());
  reverse(right.begin(),right.end());
  bool p=true;
  int curl=0;
  rep(i,0,left.size()){
    if(curl+left[i].first<0){
      p=false;
    }
    curl+=left[i].second;
  }
  int curr=0;
  rep(i,0,right.size()){
    if(curr+right[i].first<0){
      p=false;
    }
    curr+=right[i].second;
  }
  if(p && curl==curr) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
}