#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 200005
#define INF 1000000005LL
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <string,string> P;

int n;
string s;
map<P,ll> a,b;

int main(){
  cin >> n >> s;
  rep(U,0,1<<n){
    string t,w;
    rep(i,0,n){
      if(U>>i & 1) t.push_back(s[i]);
      else w.push_back(s[i]);
    }
    a[P(t,w)]++;
  }
  rep(U,0,1<<n){
    string t,w;
    rep(i,0,n){
      if(U>>i & 1) t.push_back(s[n*2-i-1]);
      else w.push_back(s[n*2-i-1]);
    }
    b[P(t,w)]++;
  }
  ll ans = 0;
  for(auto i : a){
    ans += i.second * b[i.first];
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 200005
#define INF 1000000005LL
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <string,string> P;

int n;
string s;
map<P,ll> a,b;

int main(){
  cin >> n >> s;
  rep(U,0,1<<n){
    string t,w;
    rep(i,0,n){
      if(U>>i & 1) t.push_back(s[i]);
      else w.push_back(s[i]);
    }
    a[P(t,w)]++;
  }
  rep(U,0,1<<n){
    string t,w;
    rep(i,0,n){
      if(U>>i & 1) t.push_back(s[n*2-i-1]);
      else w.push_back(s[n*2-i-1]);
    }
    b[P(t,w)]++;
  }
  ll ans = 0;
  for(auto i : a){
    ans += i.second * b[i.first];
  }
  cout << ans << endl;
  return 0;
}

```