#include <iostream>
#include <map>
using namespace std;

int main()
{
    int n;
    string s;
    cin >> n >> s;
    map<pair<string, string>, int> left;
    
    for(int mask = 0; mask < 1<<n; mask++){
        string a, b;
        for(int i=0; i<n; i++){
            if(mask&(1<<i)){
                a += s[i];
            }else{
                b += s[i];
            }
        }
        left[{a,b}]++;
    }
    long long ans = 0;
    for(int mask = 0; mask < 1<<n; mask++){
        string a, b;
        for(int i=0; i<n; i++){
            if(mask&(1<<i)){
                a += s[2*n-1-i];
            }else{
                b += s[2*n-1-i];
            }
        }
        ans += left[{a,b}];
    }
    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 <iostream>
#include <map>
using namespace std;

int main()
{
    int n;
    string s;
    cin >> n >> s;
    map<pair<string, string>, int> left;
    
    for(int mask = 0; mask < 1<<n; mask++){
        string a, b;
        for(int i=0; i<n; i++){
            if(mask&(1<<i)){
                a += s[i];
            }else{
                b += s[i];
            }
        }
        left[{a,b}]++;
    }
    long long ans = 0;
    for(int mask = 0; mask < 1<<n; mask++){
        string a, b;
        for(int i=0; i<n; i++){
            if(mask&(1<<i)){
                a += s[2*n-1-i];
            }else{
                b += s[2*n-1-i];
            }
        }
        ans += left[{a,b}];
    }
    cout << ans << endl;
    return 0;
}
```