#include <bits/stdc++.h>
#include <assert.h>

using namespace std;

int main()
{
  int N;
  string S;
  cin >> N >> S;
  map<string, long long> m;
  for (int bit = 0; bit < (1 << N); bit++)
  {
    string r, b;
    for (int i = 0; i < N; i++)
      if (bit & (1 << i))
        r += S[i];
      else
        b += S[i];
    m[r + "-" + b]++;
  }
  long long ans = 0;
  for (int bit = 0; bit < (1 << N); bit++)
  {
    string r, b;
    for (int i = 0; i < N; i++)
      if (bit & (1 << i))
        r += S[N + i];
      else
        b += S[N + i];
    reverse(r.begin(), r.end());
    reverse(b.begin(), b.end());
    ans += m[b + "-" + r];
  }
  cout << ans << endl;

  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#include <assert.h>

using namespace std;

int main()
{
  int N;
  string S;
  cin >> N >> S;
  map<string, long long> m;
  for (int bit = 0; bit < (1 << N); bit++)
  {
    string r, b;
    for (int i = 0; i < N; i++)
      if (bit & (1 << i))
        r += S[i];
      else
        b += S[i];
    m[r + "-" + b]++;
  }
  long long ans = 0;
  for (int bit = 0; bit < (1 << N); bit++)
  {
    string r, b;
    for (int i = 0; i < N; i++)
      if (bit & (1 << i))
        r += S[N + i];
      else
        b += S[N + i];
    reverse(r.begin(), r.end());
    reverse(b.begin(), b.end());
    ans += m[b + "-" + r];
  }
  cout << ans << endl;

  return 0;
}

```