#include <bits/stdc++.h>
using namespace std;
#define int long long

signed main() {
  int N;
  string s;
  cin >> N >> s;

  map<pair<string, string>, int> mp;
  for (int bit = 0; bit < (1 << N); bit++) {
    string r = "", b = "";
    for (int i = 0; i < N; i++) {
      if ((bit >> i) & 1) {
        r += s[i];
      } else {
        b += s[i];
      }
    }
    mp[{r, b}]++;
  }

  int ans = 0;
  for (int bit = 0; bit < (1 << N); bit++) {
    string r = "", b = "";
    for (int i = 0; i < N; i++) {
      if ((bit >> i) & 1) {
        r += s[N + i];
      } else {
        b += s[N + i];
      }
    }
    reverse(r.begin(), r.end());
    reverse(b.begin(), b.end());

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

  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>
using namespace std;
#define int long long

signed main() {
  int N;
  string s;
  cin >> N >> s;

  map<pair<string, string>, int> mp;
  for (int bit = 0; bit < (1 << N); bit++) {
    string r = "", b = "";
    for (int i = 0; i < N; i++) {
      if ((bit >> i) & 1) {
        r += s[i];
      } else {
        b += s[i];
      }
    }
    mp[{r, b}]++;
  }

  int ans = 0;
  for (int bit = 0; bit < (1 << N); bit++) {
    string r = "", b = "";
    for (int i = 0; i < N; i++) {
      if ((bit >> i) & 1) {
        r += s[N + i];
      } else {
        b += s[N + i];
      }
    }
    reverse(r.begin(), r.end());
    reverse(b.begin(), b.end());

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

  cout << ans << endl;
  return 0;
}

```