#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  string s;
  cin >> s;
  map<pair<string, string>, int> mp;
  for (int mask = 0; mask < (1 << n); mask++) {
    string t, r;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) t += s[i];
      else r += s[i];
    }
    reverse(r.begin(), r.end());
    mp[{t, r}]++;
  }
  long long ans = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    string t, r;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) t += s[i + n];
      else r += s[i + n];
    }
    reverse(t.begin(), t.end());
    if (mp.count({t, r})) ans += mp[{t, r}];
  }
  cout << ans << '\n';
  return 0;
}
### 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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  string s;
  cin >> s;
  map<pair<string, string>, int> mp;
  for (int mask = 0; mask < (1 << n); mask++) {
    string t, r;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) t += s[i];
      else r += s[i];
    }
    reverse(r.begin(), r.end());
    mp[{t, r}]++;
  }
  long long ans = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    string t, r;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) t += s[i + n];
      else r += s[i + n];
    }
    reverse(t.begin(), t.end());
    if (mp.count({t, r})) ans += mp[{t, r}];
  }
  cout << ans << '\n';
  return 0;
}
```