#include<iostream>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
#define rep0(i,n) for(int i=0;i<(int) n;i++)
typedef long long ll;
int n;
string s;
ll m;
map<string,ll> mp;
string ret,res;

int main (){
  cin>>n>>s;
  m=(1<<n);
  rep0(i,m){
    ret={};
    res={};
    rep0(j,n){
      if(i&(1<<j)) ret+=s[j];
      else res+=s[j];
    }
    if(res.size()) reverse(res.begin(),res.end());
    mp[ret+'+'+res]++;
  }
  ll ans=0;
  rep0(i,m){
    ret={};
    res={};
    rep0(j,n){
      if(i&(1<<j)) ret+=s[j+n];
      else res+=s[j+n];
    }
    if(ret.size()) reverse(ret.begin(),ret.end());
    ans+=mp[ret+'+'+res];
  }
  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<iostream>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
#define rep0(i,n) for(int i=0;i<(int) n;i++)
typedef long long ll;
int n;
string s;
ll m;
map<string,ll> mp;
string ret,res;

int main (){
  cin>>n>>s;
  m=(1<<n);
  rep0(i,m){
    ret={};
    res={};
    rep0(j,n){
      if(i&(1<<j)) ret+=s[j];
      else res+=s[j];
    }
    if(res.size()) reverse(res.begin(),res.end());
    mp[ret+'+'+res]++;
  }
  ll ans=0;
  rep0(i,m){
    ret={};
    res={};
    rep0(j,n){
      if(i&(1<<j)) ret+=s[j+n];
      else res+=s[j+n];
    }
    if(ret.size()) reverse(ret.begin(),ret.end());
    ans+=mp[ret+'+'+res];
  }
  cout<<ans<<endl;
  
  return 0;
}
```