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

using ll = long long int;

int n;
string str;

//map<pair<string, string>, ll> cnt;

unordered_map<string, ll> cnt;

ll ans;

int main(void){
  scanf("%d", &n);
  cin >> str;

  for (int i = 0; i < (1 << n); ++i) {
    string s1, s2;
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) s1.push_back(str[j]);
      else s2.push_back(str[j]);
    }
    cnt[(s1 + "_" + s2)] += 1;
  }

  reverse(str.begin(), str.end());


  for (int i = 0; i < (1 << n); ++i) {
    string s1, s2;
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) s1.push_back(str[j]);
      else s2.push_back(str[j]);
    }
    ans += cnt[(s1 + "_" + s2)];
  }

  printf("%lld\n", ans);

  return 0;
}
### Prompt

Please formulate 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
#include <bits/stdc++.h>
using namespace std;

using ll = long long int;

int n;
string str;

//map<pair<string, string>, ll> cnt;

unordered_map<string, ll> cnt;

ll ans;

int main(void){
  scanf("%d", &n);
  cin >> str;

  for (int i = 0; i < (1 << n); ++i) {
    string s1, s2;
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) s1.push_back(str[j]);
      else s2.push_back(str[j]);
    }
    cnt[(s1 + "_" + s2)] += 1;
  }

  reverse(str.begin(), str.end());


  for (int i = 0; i < (1 << n); ++i) {
    string s1, s2;
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) s1.push_back(str[j]);
      else s2.push_back(str[j]);
    }
    ans += cnt[(s1 + "_" + s2)];
  }

  printf("%lld\n", ans);

  return 0;
}
```