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

int main(){
  int N;
  string S;
  cin >> N >> S;
  string L = S.substr(0, N);
  string R = S.substr(N, 2 * N);
  reverse(R.begin(), R.end());
  map<pair<string, string>, ll> mp;
  ll ans = 0;
  rep(i, 0, 1 << N) {
    string r = "";
    string b = "";
    rep(j, 0, N) {
      if(i & (1 << j)) r += L[j];
      else b += L[j];
    }
    mp[{r, b}]++;
  }
  rep(i, 0, 1 << N) {
    string r = "";
    string b = "";
    rep(j, 0, N) {
      if(i & (1 << j)) r += R[j];
      else b += R[j];
    }
    ans += mp[{r, b}];
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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 rep(i, m, n) for (int i = m; i < (n); ++i)
#define rem(i, m, n) for (int i = m; i >= (n); --i)
typedef long long ll;

int main(){
  int N;
  string S;
  cin >> N >> S;
  string L = S.substr(0, N);
  string R = S.substr(N, 2 * N);
  reverse(R.begin(), R.end());
  map<pair<string, string>, ll> mp;
  ll ans = 0;
  rep(i, 0, 1 << N) {
    string r = "";
    string b = "";
    rep(j, 0, N) {
      if(i & (1 << j)) r += L[j];
      else b += L[j];
    }
    mp[{r, b}]++;
  }
  rep(i, 0, 1 << N) {
    string r = "";
    string b = "";
    rep(j, 0, N) {
      if(i & (1 << j)) r += R[j];
      else b += R[j];
    }
    ans += mp[{r, b}];
  }
  cout << ans << endl;
  return 0;
}

```