#include <bits/stdc++.h>
using namespace std;
using P = pair<string, string>;
using ll = long long;

int main() {
  int N; cin >> N;
  string S; cin >> S;
  map<P, int> mp;
  for (int i = 0; i < 1 << N; ++i) {
    string s1, s2;
    for (int j = 0; j < N; ++j) {
      ( (i >> j & 1) ? s1 : s2 ) += S[j];
    }
    ++mp[P{s1, s2}];
  }
  ll ans = 0;
  for (int i = 0; i < 1 << N; ++i) {
    string s1, s2;
    for (int j = 0; j < N; ++j) {
      ( (i >> j & 1) ? s2 : s1 ) += S[N + j];
    }
    reverse(begin(s1), end(s1));
    reverse(begin(s2), end(s2));
    ans += mp[P{s1, s2}];
  }
  cout << ans << endl;
  return 0;
}

### 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;
using P = pair<string, string>;
using ll = long long;

int main() {
  int N; cin >> N;
  string S; cin >> S;
  map<P, int> mp;
  for (int i = 0; i < 1 << N; ++i) {
    string s1, s2;
    for (int j = 0; j < N; ++j) {
      ( (i >> j & 1) ? s1 : s2 ) += S[j];
    }
    ++mp[P{s1, s2}];
  }
  ll ans = 0;
  for (int i = 0; i < 1 << N; ++i) {
    string s1, s2;
    for (int j = 0; j < N; ++j) {
      ( (i >> j & 1) ? s2 : s1 ) += S[N + j];
    }
    reverse(begin(s1), end(s1));
    reverse(begin(s2), end(s2));
    ans += mp[P{s1, s2}];
  }
  cout << ans << endl;
  return 0;
}

```