#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
using P = pair<string,string>;

int main(){
  int n;
  cin >> n;
  string s;
  cin >> s;
  string l = "", r = "";
  rep(i,n) {
    l += s[i];
    r += s[2*n-1-i];
  }
  map<P,ll> m;
  ll ans = 0;
  rep(i,1<<n) {
    string a = "", b = "";
    rep(j,n) {
      if(i & (1<<j)) a += l[j];
      else b += l[j];
    }
    m[P(a,b)]++;
  }
  rep(i,1<<n) {
    string a = "", b = "";
    rep(j,n) {
      if(i & (1<<j)) a += r[j];
      else b += r[j];
    }
    ans += m[P(a,b)];
  }
  cout << ans << endl;
  return 0;
}
### Prompt

Please provide a Cpp coded solution 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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
using P = pair<string,string>;

int main(){
  int n;
  cin >> n;
  string s;
  cin >> s;
  string l = "", r = "";
  rep(i,n) {
    l += s[i];
    r += s[2*n-1-i];
  }
  map<P,ll> m;
  ll ans = 0;
  rep(i,1<<n) {
    string a = "", b = "";
    rep(j,n) {
      if(i & (1<<j)) a += l[j];
      else b += l[j];
    }
    m[P(a,b)]++;
  }
  rep(i,1<<n) {
    string a = "", b = "";
    rep(j,n) {
      if(i & (1<<j)) a += r[j];
      else b += r[j];
    }
    ans += m[P(a,b)];
  }
  cout << ans << endl;
  return 0;
}
```