#include<bits/stdc++.h>
using namespace std;
#define int long long
map<string,map<string,int> > ma;
signed main(){
  int n;
  string s;
  cin>>n>>s;
  string s2;
  for(int i=0;i<n;i++)s2+=s[i+n];
  for(int i=0;i<(1<<n);i++){
    string t1,t2;
    for(int j=0;j<n;j++){
      if(i>>j&1)t1+=s2[j];
      else t2+=s2[j];
    }
    reverse(t1.begin(),t1.end());
    reverse(t2.begin(),t2.end());
    ma[t1][t2]++;
  }
  int ans=0;

  for(int i=0;i<(1<<n);i++){
    string t1,t2;
    for(int j=0;j<n;j++){
      if(i>>j&1)t1+=s[j];
      else t2+=s[j];
    }
    ans+=ma[t1][t2];
  }
  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;
#define int long long
map<string,map<string,int> > ma;
signed main(){
  int n;
  string s;
  cin>>n>>s;
  string s2;
  for(int i=0;i<n;i++)s2+=s[i+n];
  for(int i=0;i<(1<<n);i++){
    string t1,t2;
    for(int j=0;j<n;j++){
      if(i>>j&1)t1+=s2[j];
      else t2+=s2[j];
    }
    reverse(t1.begin(),t1.end());
    reverse(t2.begin(),t2.end());
    ma[t1][t2]++;
  }
  int ans=0;

  for(int i=0;i<(1<<n);i++){
    string t1,t2;
    for(int j=0;j<n;j++){
      if(i>>j&1)t1+=s[j];
      else t2+=s[j];
    }
    ans+=ma[t1][t2];
  }
  cout<<ans<<endl;
  return 0;
}

```