#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

int main () {
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n;
  string s;
  cin >> n >> s;
  map < pair < string, string >, int > mp;

  for (int i = 0; i < (1 << n); i++) {
    string r, b;
    for (int j = 0; j < n; j++) {
      if ((i >> j) & 1) r += s[j];
      else b += s[j];
    }
    mp[{r, b}]++;
  }

  ll ans = 0;
  for (int i = 0; i < (1 << n); i++) {
    string r, b;
    for (int j = n; j < 2 * n; j++) {
      if ((i >> (j - n)) & 1) b += s[j];
      else r += s[j];
    }
    reverse (r.begin(), r.end());
    reverse (b.begin(), b.end());

    ans += mp[{r, b}];
  }

  cout << ans << '\n';
}

### Prompt

Develop a solution in cpp to the problem described below:

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 int ll;

int main () {
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n;
  string s;
  cin >> n >> s;
  map < pair < string, string >, int > mp;

  for (int i = 0; i < (1 << n); i++) {
    string r, b;
    for (int j = 0; j < n; j++) {
      if ((i >> j) & 1) r += s[j];
      else b += s[j];
    }
    mp[{r, b}]++;
  }

  ll ans = 0;
  for (int i = 0; i < (1 << n); i++) {
    string r, b;
    for (int j = n; j < 2 * n; j++) {
      if ((i >> (j - n)) & 1) b += s[j];
      else r += s[j];
    }
    reverse (r.begin(), r.end());
    reverse (b.begin(), b.end());

    ans += mp[{r, b}];
  }

  cout << ans << '\n';
}

```