#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <utility>

using namespace std;

int main()
{
    int N;
    cin>>N;
    string S;
    cin>>S;

    map<pair<string,string>, int> M;

    for (int b=0; b<1<<N; b++)
    {
        string sr;
        string sb;
        for (int i=0; i<N; i++)
            (b>>i&1 ? sr : sb) += S[i];
        M[make_pair(sr,sb)]++;
    }

    long long ans = 0;
    for (int b=0; b<1<<N; b++)
    {
        string sr;
        string sb;
        for (int i=0; i<N; i++)
            (b>>i&1 ? sr : sb) += S[2*N-i-1];
        ans += M[make_pair(sr,sb)];
    }

    cout<<ans<<endl;
}

### Prompt

Develop a solution in Cpp 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 <iostream>
#include <vector>
#include <map>
#include <string>
#include <utility>

using namespace std;

int main()
{
    int N;
    cin>>N;
    string S;
    cin>>S;

    map<pair<string,string>, int> M;

    for (int b=0; b<1<<N; b++)
    {
        string sr;
        string sb;
        for (int i=0; i<N; i++)
            (b>>i&1 ? sr : sb) += S[i];
        M[make_pair(sr,sb)]++;
    }

    long long ans = 0;
    for (int b=0; b<1<<N; b++)
    {
        string sr;
        string sb;
        for (int i=0; i<N; i++)
            (b>>i&1 ? sr : sb) += S[2*N-i-1];
        ans += M[make_pair(sr,sb)];
    }

    cout<<ans<<endl;
}

```