#include<bits/stdc++.h>
using namespace std;
#define int long long
map<pair<string, string>, int> mp;
int n, rt;
string s;
int32_t main(){
    cin >> n >> s;
    for(int i=0;i<(1<<n);i++){
        string a = "", b = "";
        for(int j=0;j<n;j++){
            if(i&(1LL<<j)) a+=s[j];
            else b+=s[j];
        }
        mp[{a, b}]++;
    }
    for(int i=0;i<(1<<n);i++){
        string a = "", b = "";
        for(int j=0;j<n;j++){
            if(i&(1LL<<j)) a+=s[j+n];
            else b+=s[j+n];
        }
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        rt+=mp[{b, a}];
    }
    cout << rt;
}
### Prompt

Please formulate a CPP solution 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>
using namespace std;
#define int long long
map<pair<string, string>, int> mp;
int n, rt;
string s;
int32_t main(){
    cin >> n >> s;
    for(int i=0;i<(1<<n);i++){
        string a = "", b = "";
        for(int j=0;j<n;j++){
            if(i&(1LL<<j)) a+=s[j];
            else b+=s[j];
        }
        mp[{a, b}]++;
    }
    for(int i=0;i<(1<<n);i++){
        string a = "", b = "";
        for(int j=0;j<n;j++){
            if(i&(1LL<<j)) a+=s[j+n];
            else b+=s[j+n];
        }
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        rt+=mp[{b, a}];
    }
    cout << rt;
}
```