#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n ;
string  s ;
cin >> n >> s ;
string f=s.substr(0,n);
string b=s.substr(n,n);
reverse(b.begin(), b.end());
map<pair<string,string>, long long > ma;
rep(i,1LL<<n){
  string re="",bl="";
  rep(j,n){
    if(i>>j&1LL)re+=f[j];
    else bl+=f[j];
  }
  pair<string,string> p={re,bl};
  ma[p]++;
}
rep(i,1LL<<n){
  string re="",bl="";
  rep(j,n){
    if(i>>j&1LL)re+=b[j];
    else bl+=b[j];
  }
  pair<string,string> p={re,bl};
  ans+=ma[p];
}
cout << ans << endl;
return 0;
}

### Prompt

Generate 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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n ;
string  s ;
cin >> n >> s ;
string f=s.substr(0,n);
string b=s.substr(n,n);
reverse(b.begin(), b.end());
map<pair<string,string>, long long > ma;
rep(i,1LL<<n){
  string re="",bl="";
  rep(j,n){
    if(i>>j&1LL)re+=f[j];
    else bl+=f[j];
  }
  pair<string,string> p={re,bl};
  ma[p]++;
}
rep(i,1LL<<n){
  string re="",bl="";
  rep(j,n){
    if(i>>j&1LL)re+=b[j];
    else bl+=b[j];
  }
  pair<string,string> p={re,bl};
  ans+=ma[p];
}
cout << ans << endl;
return 0;
}

```