#include <bits/stdc++.h>
#define ll long long int
using namespace std;
const ll NMAX = 1e5+7;
std::map<pair<string ,string>, int> ma;
ll n, res=0;
string s;
int main(){
   std::cin >> n >> s;
   for(int i = 0; i<(1<<n); i++){
     string a = "", b = "";
     for(int j = 0; j<n; j++){
      if(i&(1<<j)) a+=s[j];
      else b+=s[j];
     }
     ma[{a,b}]++;
   }

   for(int i = 0; i<(1<<n); i++){
     string a = "", b = "";
     for(int j = 0; j<n; j++){
      if(i&(1<<j)) a+=s[n+j];
      else b+=s[n+j];
     }
     reverse(a.begin(), a.end());
     reverse(b.begin(), b.end());
     res+=ma[{b,a}];
   }
   std::cout << res << '\n';
}

### 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>
#define ll long long int
using namespace std;
const ll NMAX = 1e5+7;
std::map<pair<string ,string>, int> ma;
ll n, res=0;
string s;
int main(){
   std::cin >> n >> s;
   for(int i = 0; i<(1<<n); i++){
     string a = "", b = "";
     for(int j = 0; j<n; j++){
      if(i&(1<<j)) a+=s[j];
      else b+=s[j];
     }
     ma[{a,b}]++;
   }

   for(int i = 0; i<(1<<n); i++){
     string a = "", b = "";
     for(int j = 0; j<n; j++){
      if(i&(1<<j)) a+=s[n+j];
      else b+=s[n+j];
     }
     reverse(a.begin(), a.end());
     reverse(b.begin(), b.end());
     res+=ma[{b,a}];
   }
   std::cout << res << '\n';
}

```