#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <map>

using namespace std; typedef long long ll;
typedef pair<string,string> P; const int INF = 1e9+1;
int main() {
  int n; cin>>n;
  string s; cin>>s;

  map<P, ll> mp;
  for(int i=0;i<(1<<n);i++) {
    string s1, s2;
    for(int j=0;j<n;j++) {
      if (i & (1<<j)) s1+=s[j];
      else s2+=s[j];
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    mp[make_pair(s1,s2)]++;
  }
  ll an=0;
  for(int i=0;i<(1<<n);i++) {
    string s1, s2;
    for(int j=0;j<n;j++) {
      if (i & (1<<j)) s1+=s[n+j];
      else s2+=s[n+j];
    }
    an += mp[make_pair(s1,s2)];
  }
  cout<<an<<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 <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <map>

using namespace std; typedef long long ll;
typedef pair<string,string> P; const int INF = 1e9+1;
int main() {
  int n; cin>>n;
  string s; cin>>s;

  map<P, ll> mp;
  for(int i=0;i<(1<<n);i++) {
    string s1, s2;
    for(int j=0;j<n;j++) {
      if (i & (1<<j)) s1+=s[j];
      else s2+=s[j];
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    mp[make_pair(s1,s2)]++;
  }
  ll an=0;
  for(int i=0;i<(1<<n);i++) {
    string s1, s2;
    for(int j=0;j<n;j++) {
      if (i & (1<<j)) s1+=s[n+j];
      else s2+=s[n+j];
    }
    an += mp[make_pair(s1,s2)];
  }
  cout<<an<<endl;
}
```