#include <bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin >> N;
  string S;
  cin >> S;
  string S1 = S.substr(0, N);
  map<pair<string, string>, int> mp;
  for (int i = 0; i < (1 << N); i++){
    string red = "";
    string blue = "";
    for (int j = 0; j < N; j++){
      if (i >> j & 1){
        red += S1[j];
      } else {
        blue += S1[j];
      }
    }
    mp[make_pair(red, blue)]++;
  }
  long long ans = 0;
  string S2 = S.substr(N);
  for (int i = 0; i < (1 << N); i++){
    string red = "";
    string blue = "";
    for (int j = N - 1; j >= 0; j--){
      if (i >> j & 1){
        red += S2[j];
      } else {
        blue += S2[j];
      }
    }
    ans += mp[make_pair(blue, red)];
  }
  cout << ans << endl;
}
### 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;
int main(){
  int N;
  cin >> N;
  string S;
  cin >> S;
  string S1 = S.substr(0, N);
  map<pair<string, string>, int> mp;
  for (int i = 0; i < (1 << N); i++){
    string red = "";
    string blue = "";
    for (int j = 0; j < N; j++){
      if (i >> j & 1){
        red += S1[j];
      } else {
        blue += S1[j];
      }
    }
    mp[make_pair(red, blue)]++;
  }
  long long ans = 0;
  string S2 = S.substr(N);
  for (int i = 0; i < (1 << N); i++){
    string red = "";
    string blue = "";
    for (int j = N - 1; j >= 0; j--){
      if (i >> j & 1){
        red += S2[j];
      } else {
        blue += S2[j];
      }
    }
    ans += mp[make_pair(blue, red)];
  }
  cout << ans << endl;
}
```