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

int main(){ 
  ll N; cin >> N;
  string s; cin >> s;
  map<pair<string,string>,ll> mp;
  ll ans = 0;
  for(ll bit = 0; bit < (1<<N); bit++){
    string a = "",b = "";
    for(ll i = 0;  i < N; i++){ 
      if((bit>>i)&1){a += s[i];}
      else{ b += s[i];}
    }
    mp[{a,b}]++;
  }
  for(ll bit = 0; bit < (1<<N); bit++){ 
    string a = "",b = "";
    for(ll i = 0; i < N; i++){ 
      if((bit>>i)&1){ a += s[i+N];}
      else{b += s[i+N];}
    }
    reverse(a.begin(),a.end()); reverse(b.begin(),b.end());
    ans += mp[{a,b}];
  }
  
  cout << ans << endl;
  return 0;
}
### 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;
typedef long long ll;

int main(){ 
  ll N; cin >> N;
  string s; cin >> s;
  map<pair<string,string>,ll> mp;
  ll ans = 0;
  for(ll bit = 0; bit < (1<<N); bit++){
    string a = "",b = "";
    for(ll i = 0;  i < N; i++){ 
      if((bit>>i)&1){a += s[i];}
      else{ b += s[i];}
    }
    mp[{a,b}]++;
  }
  for(ll bit = 0; bit < (1<<N); bit++){ 
    string a = "",b = "";
    for(ll i = 0; i < N; i++){ 
      if((bit>>i)&1){ a += s[i+N];}
      else{b += s[i+N];}
    }
    reverse(a.begin(),a.end()); reverse(b.begin(),b.end());
    ans += mp[{a,b}];
  }
  
  cout << ans << endl;
  return 0;
}
```