#include "bits/stdc++.h"

using namespace std;

typedef long long ll;

int main() {
  ll N;
  cin >> N;

  string S;
  cin >> S;

  string s1 = S.substr(0, N);
  string s2 = S.substr(N, 2 * N);
  reverse(s2.begin(), s2.end());

  unordered_map<string, ll> m1, m2;
  for (ll i = 0; i < (1ll << N); i++) {
    pair<string, string> p1, p2;

    for (ll j = 0; j < N; j++) {
      if (i & (1ll << j)) {
        p1.first += s1[j];
        p2.first += s2[j];
      } else {
        p1.second += s1[j];
        p2.second += s2[j];
      }
    }


    m1[p1.first + "," + p1.second] += 1;
    m2[p2.first + "," + p2.second] += 1;
  }

  ll ans = 0;
  for (auto p: m1) {
    ans += p.second * m2[p.first];
  }

  cout << ans << endl;
}

### 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;

typedef long long ll;

int main() {
  ll N;
  cin >> N;

  string S;
  cin >> S;

  string s1 = S.substr(0, N);
  string s2 = S.substr(N, 2 * N);
  reverse(s2.begin(), s2.end());

  unordered_map<string, ll> m1, m2;
  for (ll i = 0; i < (1ll << N); i++) {
    pair<string, string> p1, p2;

    for (ll j = 0; j < N; j++) {
      if (i & (1ll << j)) {
        p1.first += s1[j];
        p2.first += s2[j];
      } else {
        p1.second += s1[j];
        p2.second += s2[j];
      }
    }


    m1[p1.first + "," + p1.second] += 1;
    m2[p2.first + "," + p2.second] += 1;
  }

  ll ans = 0;
  for (auto p: m1) {
    ans += p.second * m2[p.first];
  }

  cout << ans << endl;
}

```