#include <bits/stdc++.h>
using namespace std;
int main(){
  long long n,ans=0;
  string s,l,r;
  cin >> n >> s;
  l=s.substr(0,n);
  r=s.substr(n,n);
  string red,blue;
  map<string, int> mp;
  for(int bit=0;bit<(1 << n);bit++){
    red="";
    blue="";
    for(int i=0;i<n;i++){
      if(bit & (1 << i)){
        red+=l[i];
      }else{
        blue+=l[i];
      }
    }
    string t=red+","+blue;
    mp[t]++;
  }
  for(int bit=0;bit<(1 << n);bit++){
    red="";
    blue="";
    for(int i=0;i<n;i++){
      if(bit & (1 << i)){
        red+=r[i];
      }else{
        blue+=r[i];
      }
    }
    reverse(red.begin(), red.end());
    reverse(blue.begin(), blue.end());
    string t=blue+","+red;
    ans+=mp[t];
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
int main(){
  long long n,ans=0;
  string s,l,r;
  cin >> n >> s;
  l=s.substr(0,n);
  r=s.substr(n,n);
  string red,blue;
  map<string, int> mp;
  for(int bit=0;bit<(1 << n);bit++){
    red="";
    blue="";
    for(int i=0;i<n;i++){
      if(bit & (1 << i)){
        red+=l[i];
      }else{
        blue+=l[i];
      }
    }
    string t=red+","+blue;
    mp[t]++;
  }
  for(int bit=0;bit<(1 << n);bit++){
    red="";
    blue="";
    for(int i=0;i<n;i++){
      if(bit & (1 << i)){
        red+=r[i];
      }else{
        blue+=r[i];
      }
    }
    reverse(red.begin(), red.end());
    reverse(blue.begin(), blue.end());
    string t=blue+","+red;
    ans+=mp[t];
  }
  cout << ans << endl;
  return 0;
}

```