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

int main(){
  int N;
  string S;
  cin>>N>>S;
  // 半分に分割し、右側は反転
  string L = S.substr(0,N);
  string R = S.substr(N,2*N);
  reverse(R.begin(),R.end());
  // 左側の塗分け方を全探索
  map<P,ll> mp;
  for(int i=0;i<(1<<N);i++){
    string red = "";
    string blue = "";
    for(int j=0;j<N;j++){
      if(i & (1<<j)) red += L[j];
      else blue += L[j];
    }
    mp[{red,blue}]++;
  }
  // 右側の塗分け方を全探索
  ll ans = 0;
  for(int i=0;i<(1<<N);i++){
    string red = "";
    string blue = "";
    for(int j=0;j<N;j++){
      if(i & (1<<j)) red += R[j];
      else blue += R[j];
    }
    ans += mp[{red,blue}];
  }
  cout<<ans<<endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
typedef pair<string,string> P;

int main(){
  int N;
  string S;
  cin>>N>>S;
  // 半分に分割し、右側は反転
  string L = S.substr(0,N);
  string R = S.substr(N,2*N);
  reverse(R.begin(),R.end());
  // 左側の塗分け方を全探索
  map<P,ll> mp;
  for(int i=0;i<(1<<N);i++){
    string red = "";
    string blue = "";
    for(int j=0;j<N;j++){
      if(i & (1<<j)) red += L[j];
      else blue += L[j];
    }
    mp[{red,blue}]++;
  }
  // 右側の塗分け方を全探索
  ll ans = 0;
  for(int i=0;i<(1<<N);i++){
    string red = "";
    string blue = "";
    for(int j=0;j<N;j++){
      if(i & (1<<j)) red += R[j];
      else blue += R[j];
    }
    ans += mp[{red,blue}];
  }
  cout<<ans<<endl;
  return 0;
}

```