#include <bits/stdc++.h>
#define ll long long
using namespace std;
int N;
char s[40],a[40],b[40],c[40],d[40];
int cnt,dnt;
ll ans;
ll f[40][40];
void dp() {
    memset(f,0,sizeof f);
    f[0][0]=1;
    for(int i=0;i<=cnt;++i)
        for(int j=0;j<=dnt;++j) if(i+j)
            f[i][j]=f[i-1][j]*(c[i]==b[i+j])+f[i][j-1]*(d[j]==b[i+j]);//,printf("f[%d][%d]=%lld\n",i,j,f[i][j]);
    ans+=f[cnt][dnt];
}
int main() {
    scanf("%d",&N);
    scanf("%s",s+1);
    for(int i=1;i<=N;++i) a[i]=s[i],b[i]=s[N+N+1-i];
    for(int i=0;i<(1<<N);++i) {
        dnt=cnt=0;
        for(int j=0;j<N;++j) {
            if((1<<j)&i) c[++cnt]=a[j+1];
            else    d[++dnt]=a[j+1];
        }
        dp();
    }
    printf("%lld\n",ans);
    return 0;
}
### 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 <bits/stdc++.h>
#define ll long long
using namespace std;
int N;
char s[40],a[40],b[40],c[40],d[40];
int cnt,dnt;
ll ans;
ll f[40][40];
void dp() {
    memset(f,0,sizeof f);
    f[0][0]=1;
    for(int i=0;i<=cnt;++i)
        for(int j=0;j<=dnt;++j) if(i+j)
            f[i][j]=f[i-1][j]*(c[i]==b[i+j])+f[i][j-1]*(d[j]==b[i+j]);//,printf("f[%d][%d]=%lld\n",i,j,f[i][j]);
    ans+=f[cnt][dnt];
}
int main() {
    scanf("%d",&N);
    scanf("%s",s+1);
    for(int i=1;i<=N;++i) a[i]=s[i],b[i]=s[N+N+1-i];
    for(int i=0;i<(1<<N);++i) {
        dnt=cnt=0;
        for(int j=0;j<N;++j) {
            if((1<<j)&i) c[++cnt]=a[j+1];
            else    d[++dnt]=a[j+1];
        }
        dp();
    }
    printf("%lld\n",ans);
    return 0;
}
```