#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int N;
  string S;
  cin>>N>>S;
  reverse(S.begin()+N,S.end());
  map<pair<string,string>,int> A,B;
  for(int i=0;i<(1<<N);i++){
    bitset<18> C(i);
    string T="",U="";
    for(int j=0;j<N;j++){
      if(C.test(j))
        T+=S[j];
      else
        U+=S[j];
    }
    A[make_pair(T,U)]++;
  }
  for(int i=0;i<(1<<N);i++){
    bitset<18> C(i);
    string T="",U="";
    for(int j=0;j<N;j++){
      if(C.test(j))
        T+=S[j+N];
      else
        U+=S[j+N];
    }
    B[make_pair(T,U)]++;
  }
  int ans=0;
  for(pair<pair<string,string>,int> p:A){
    ans+=p.second*B[p.first];
  }
  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 int long long
signed main(){
  int N;
  string S;
  cin>>N>>S;
  reverse(S.begin()+N,S.end());
  map<pair<string,string>,int> A,B;
  for(int i=0;i<(1<<N);i++){
    bitset<18> C(i);
    string T="",U="";
    for(int j=0;j<N;j++){
      if(C.test(j))
        T+=S[j];
      else
        U+=S[j];
    }
    A[make_pair(T,U)]++;
  }
  for(int i=0;i<(1<<N);i++){
    bitset<18> C(i);
    string T="",U="";
    for(int j=0;j<N;j++){
      if(C.test(j))
        T+=S[j+N];
      else
        U+=S[j+N];
    }
    B[make_pair(T,U)]++;
  }
  int ans=0;
  for(pair<pair<string,string>,int> p:A){
    ans+=p.second*B[p.first];
  }
  cout<<ans<<endl;
}
```