#include<stdio.h>
#include<string.h>
#define N 55
#define LL long long
LL s[N][N];
char a[N],b[N],c[N];

int main(){
	int n;
	LL ans;
	scanf("%d%s",&n,a);
	ans=0;
	for(int t=0;t<(1<<n);++t){
		int k1=0,k2=0;
		for(int j=0;j<n;++j){
			if(t&1<<j)b[k1++]=a[j];
			else c[k2++]=a[j];
        }
		b[k1]=c[k2]='\0';
		memset(s,0,sizeof(s));
        s[0][0]=1;
		for(int i=0;i<=k1;++i){
			for(int j=0;j<=k2;++j){
				if(a[n*2-1-i-j]==b[i])
				s[i+1][j]+=s[i][j];
				if(a[n*2-1-i-j]==c[j])
				s[i][j+1]+=s[i][j];
			}
		}
		ans+=s[k1][k2];
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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<stdio.h>
#include<string.h>
#define N 55
#define LL long long
LL s[N][N];
char a[N],b[N],c[N];

int main(){
	int n;
	LL ans;
	scanf("%d%s",&n,a);
	ans=0;
	for(int t=0;t<(1<<n);++t){
		int k1=0,k2=0;
		for(int j=0;j<n;++j){
			if(t&1<<j)b[k1++]=a[j];
			else c[k2++]=a[j];
        }
		b[k1]=c[k2]='\0';
		memset(s,0,sizeof(s));
        s[0][0]=1;
		for(int i=0;i<=k1;++i){
			for(int j=0;j<=k2;++j){
				if(a[n*2-1-i-j]==b[i])
				s[i+1][j]+=s[i][j];
				if(a[n*2-1-i-j]==c[j])
				s[i][j+1]+=s[i][j];
			}
		}
		ans+=s[k1][k2];
	}
	printf("%lld\n",ans);
	return 0;
}
```