#include <iostream>
#include <vector>
#include <map>
using namespace std;

int main(){
  int N;
  string S;
  cin >> N >> S;
  vector<map<string,int>> mp(N+1);
  for(int i = 0; i < (1<<N); ++i){
    string A(N,'?');
    int a = 0, b = 0;
    for(int j = 0; j < N; ++j){
      if((i>>j)&1) A[a++] = S[j];
      else A[N-1-b++] = S[j];
    }
    ++mp[a][A];
  }
  long long ans = 0;
  for(int i = 0; i < (1<<N); ++i){
    string A(N,'?');
    int a = 0, b = 0;
    for(int j = 0; j < N; ++j){
      if((i>>j)&1) A[a++] = S[2*N-1-j];
      else A[N-1-b++] = S[2*N-1-j];
    }
    ans += mp[a][A];
  }
  cout << ans << endl;
}

### 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 <iostream>
#include <vector>
#include <map>
using namespace std;

int main(){
  int N;
  string S;
  cin >> N >> S;
  vector<map<string,int>> mp(N+1);
  for(int i = 0; i < (1<<N); ++i){
    string A(N,'?');
    int a = 0, b = 0;
    for(int j = 0; j < N; ++j){
      if((i>>j)&1) A[a++] = S[j];
      else A[N-1-b++] = S[j];
    }
    ++mp[a][A];
  }
  long long ans = 0;
  for(int i = 0; i < (1<<N); ++i){
    string A(N,'?');
    int a = 0, b = 0;
    for(int j = 0; j < N; ++j){
      if((i>>j)&1) A[a++] = S[2*N-1-j];
      else A[N-1-b++] = S[2*N-1-j];
    }
    ans += mp[a][A];
  }
  cout << ans << endl;
}

```