#include <iostream>
#include <map>
using namespace std;
typedef pair<string, string> P;
typedef long long ll;
map<P, ll> p,q;
int main(void){
    int n;
    ll m=0;
    string s;
    cin>>n>>s;
    for(int i=0;i<(1<<n);i++){
        string a="",b="",c="",d="";
        for(int j=0;j<n;j++){
            if(((i>>j)&1)==1){
                a+=s[j];
                c+=s[n*2-1-j];
            }else{
                b+=s[j];
                d+=s[n*2-1-j];
            }
        }
        p[P(a, b)]++;
        q[P(c, d)]++;
    }
    for(auto x=p.begin();x!=p.end();x++){
        m+=(x->second)*q[(x->first)];
    }
    cout<<m<<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 <map>
using namespace std;
typedef pair<string, string> P;
typedef long long ll;
map<P, ll> p,q;
int main(void){
    int n;
    ll m=0;
    string s;
    cin>>n>>s;
    for(int i=0;i<(1<<n);i++){
        string a="",b="",c="",d="";
        for(int j=0;j<n;j++){
            if(((i>>j)&1)==1){
                a+=s[j];
                c+=s[n*2-1-j];
            }else{
                b+=s[j];
                d+=s[n*2-1-j];
            }
        }
        p[P(a, b)]++;
        q[P(c, d)]++;
    }
    for(auto x=p.begin();x!=p.end();x++){
        m+=(x->second)*q[(x->first)];
    }
    cout<<m<<endl;
}

```