#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,i,j,mx,t,x,ans,f[19][19],a[19],b[19];
char s[101];
int main(){
	scanf("%lld",&n);
	scanf("%s",s+1);
	mx=1ll<<n;mx--;
	for(t=0;t<=mx;t++){
		a[0]=b[0]=0;
		for(i=n*2;i>n;i--){
			x=n*2-i;
			if(t&(1ll<<x))a[++a[0]]=s[i]-'a';
			 else b[++b[0]]=s[i]-'a';
		}
		memset(f,0,sizeof(f));
		f[0][0]=1;
		for(i=1;i<=n;i++){
			for(j=0;j<=min(a[0],i-1);j++){
				if(i-1-j>b[0])continue;
				if(j<a[0]&&s[i]-'a'==a[j+1])f[i][j+1]+=f[i-1][j];
				if(i-j<=b[0]&&s[i]-'a'==b[i-j])f[i][j]+=f[i-1][j];
			}
		}
		ans+=f[n][a[0]];
	}
	printf("%lld",ans);
}
### 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<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,i,j,mx,t,x,ans,f[19][19],a[19],b[19];
char s[101];
int main(){
	scanf("%lld",&n);
	scanf("%s",s+1);
	mx=1ll<<n;mx--;
	for(t=0;t<=mx;t++){
		a[0]=b[0]=0;
		for(i=n*2;i>n;i--){
			x=n*2-i;
			if(t&(1ll<<x))a[++a[0]]=s[i]-'a';
			 else b[++b[0]]=s[i]-'a';
		}
		memset(f,0,sizeof(f));
		f[0][0]=1;
		for(i=1;i<=n;i++){
			for(j=0;j<=min(a[0],i-1);j++){
				if(i-1-j>b[0])continue;
				if(j<a[0]&&s[i]-'a'==a[j+1])f[i][j+1]+=f[i-1][j];
				if(i-j<=b[0]&&s[i]-'a'==b[i-j])f[i][j]+=f[i-1][j];
			}
		}
		ans+=f[n][a[0]];
	}
	printf("%lld",ans);
}
```