#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<map>
using namespace std;

int n;
long long ans;
string s;
map<pair<string,string>,int> cnt;

void pre_dfs(string s1,string s2,int x){
    if(x==n){
        cnt[make_pair(s1,s2)]++;
        return;
    }
    pre_dfs(s1+s[x],s2,x+1);
    pre_dfs(s1,s2+s[x],x+1);
}

void dfs(string s1,string s2,int x){
    if(x==n-1){
        ans+=cnt[make_pair(s2,s1)];
        return ;
    }
    dfs(s1+s[x],s2,x-1);
    dfs(s1,s2+s[x],x-1);
}

int main(){
    scanf("%d",&n);
    cin >> s;
    string s1,s2;
    pre_dfs(s1,s2,0);
    dfs(s1,s2,2*n-1);
    printf("%lld",ans);
}

### Prompt

Create a solution in Cpp for 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<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<map>
using namespace std;

int n;
long long ans;
string s;
map<pair<string,string>,int> cnt;

void pre_dfs(string s1,string s2,int x){
    if(x==n){
        cnt[make_pair(s1,s2)]++;
        return;
    }
    pre_dfs(s1+s[x],s2,x+1);
    pre_dfs(s1,s2+s[x],x+1);
}

void dfs(string s1,string s2,int x){
    if(x==n-1){
        ans+=cnt[make_pair(s2,s1)];
        return ;
    }
    dfs(s1+s[x],s2,x-1);
    dfs(s1,s2+s[x],x-1);
}

int main(){
    scanf("%d",&n);
    cin >> s;
    string s1,s2;
    pre_dfs(s1,s2,0);
    dfs(s1,s2,2*n-1);
    printf("%lld",ans);
}

```