#include <bits/stdc++.h>
using namespace std;
#define ll long long

map<pair<string,string>,ll>M;
int main(){
  int n;cin>>n;
  string s;cin>>s;
  for(int i=0;i<(1<<n);++i){
    string red="",blue="";
    for(int j=0;j<n;++j){
      if(i&(1<<j))red+=s[j];
      else blue+=s[j];
    }
    ++M[make_pair(red,blue)];
  }
  ll ans=0;
  for(int i=0;i<(1<<n);++i){
    string red="",blue="";
    for(int j=0;j<n;++j){
      if(i&(1<<j))red+=s[j+n];
      else blue+=s[j+n];
    }
    reverse(red.begin(),red.end());
    reverse(blue.begin(),blue.end());
    ans+=M[make_pair(blue,red)];
  }
  cout<<ans<<endl;
}
### Prompt

Create a solution in CPP for the following problem:

You are given a string S of length 2N consisting of lowercase English letters.

There are 2^{2N} ways to color each character in S red or blue. Among these ways, how many satisfy the following condition?

* The string obtained by reading the characters painted red from left to right is equal to the string obtained by reading the characters painted blue from right to left.

Constraints

* 1 \leq N \leq 18
* The length of S is 2N.
* S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the number of ways to paint the string that satisfy the condition.

Examples

Input

4
cabaacba


Output

4


Input

11
mippiisssisssiipsspiim


Output

504


Input

4
abcdefgh


Output

0


Input

18
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa


Output

9075135300

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define ll long long

map<pair<string,string>,ll>M;
int main(){
  int n;cin>>n;
  string s;cin>>s;
  for(int i=0;i<(1<<n);++i){
    string red="",blue="";
    for(int j=0;j<n;++j){
      if(i&(1<<j))red+=s[j];
      else blue+=s[j];
    }
    ++M[make_pair(red,blue)];
  }
  ll ans=0;
  for(int i=0;i<(1<<n);++i){
    string red="",blue="";
    for(int j=0;j<n;++j){
      if(i&(1<<j))red+=s[j+n];
      else blue+=s[j+n];
    }
    reverse(red.begin(),red.end());
    reverse(blue.begin(),blue.end());
    ans+=M[make_pair(blue,red)];
  }
  cout<<ans<<endl;
}
```