#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int(i)=0;i<(n);i++)

int main(){
  int n,ans=1;string s;cin>>n>>s;
  if(n%2) ans=0;
  else REP(i,n/2) if(s.at(i)!=s.at(i+n/2)) ans=0;
  cout<<(ans?"Yes":"No");
}
### Prompt

Generate a CPP solution to the following problem:

Given are a positive integer N and a string S of length N consisting of lowercase English letters.

Determine whether the string is a concatenation of two copies of some string. That is, determine whether there is a string T such that S = T + T.

Constraints

* 1 \leq N \leq 100
* S consists of lowercase English letters.
* |S| = N

Input

Input is given from Standard Input in the following format:


N
S


Output

If S is a concatenation of two copies of some string, print `Yes`; otherwise, print `No`.

Examples

Input

6
abcabc


Output

Yes


Input

6
abcadc


Output

No


Input

1
z


Output

No

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int(i)=0;i<(n);i++)

int main(){
  int n,ans=1;string s;cin>>n>>s;
  if(n%2) ans=0;
  else REP(i,n/2) if(s.at(i)!=s.at(i+n/2)) ans=0;
  cout<<(ans?"Yes":"No");
}
```