#include<cstdio>
#include<algorithm>
#define LL long long 
int lena,lenb,a[100],b[100],n;
char s[50];
LL ans,f[50][50];
using namespace std;
void pan()
{
	f[n][1]=(s[b[lenb]]==s[n]);
	f[n][0]=(s[a[lena]]==s[n]);
	for (int i=n+1;i<n*2;i++)
	  for (int j=0;j<=min(i-n+1,lenb);j++)
	  {
	  	f[i][j]=0;
	  	if (lena>0)
	  	if (j<i-n+1)
	  	if (s[a[lena-i+n+j]]==s[i]) f[i][j]+=f[i-1][j];
	  	if (j>=1)
	  	if (lenb>0)
	  	if (s[i]==s[b[lenb-j+1]]) f[i][j]+=f[i-1][j-1];
	  }
	ans+=f[n*2-1][lenb]; 
}
void dfs(int x)
{
	if (x>=n)
	{
		pan(); 
	}
	else
	{
		lena++; a[lena]=x; dfs(x+1); lena--;
		lenb++; b[lenb]=x; dfs(x+1); lenb--; 
	}
}
int main()
{
	scanf("%d",&n);
	scanf("%s",s);
    dfs(0);
    printf("%lld",ans);
}
### Prompt

Develop a solution in Cpp to the problem described below:

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<algorithm>
#define LL long long 
int lena,lenb,a[100],b[100],n;
char s[50];
LL ans,f[50][50];
using namespace std;
void pan()
{
	f[n][1]=(s[b[lenb]]==s[n]);
	f[n][0]=(s[a[lena]]==s[n]);
	for (int i=n+1;i<n*2;i++)
	  for (int j=0;j<=min(i-n+1,lenb);j++)
	  {
	  	f[i][j]=0;
	  	if (lena>0)
	  	if (j<i-n+1)
	  	if (s[a[lena-i+n+j]]==s[i]) f[i][j]+=f[i-1][j];
	  	if (j>=1)
	  	if (lenb>0)
	  	if (s[i]==s[b[lenb-j+1]]) f[i][j]+=f[i-1][j-1];
	  }
	ans+=f[n*2-1][lenb]; 
}
void dfs(int x)
{
	if (x>=n)
	{
		pan(); 
	}
	else
	{
		lena++; a[lena]=x; dfs(x+1); lena--;
		lenb++; b[lenb]=x; dfs(x+1); lenb--; 
	}
}
int main()
{
	scanf("%d",&n);
	scanf("%s",s);
    dfs(0);
    printf("%lld",ans);
}
```