#include<bits/stdc++.h>
using namespace std;
const int N=233,M=20;
char a[N],x[M],y[M];
int myxn,myyn,n;
long long dp[M][M];

long long solve(int i)
{
	if(i==n)
	{
		//printf("233\n");
		memset(dp,0,sizeof(dp));
		dp[0][myxn]=1;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<=myxn;j++)if(dp[i][j])
			{
				if(j && a[i+n]==x[j-1])dp[i+1][j-1]+=dp[i][j];
				int k=myxn-j-i+myyn;
				//printf("i=%d j=%d k=%d\n",i,j,k);
				if(k && a[i+n]==y[k-1])dp[i+1][j]+=dp[i][j];
			}
		}
		return dp[n][0];
	}
	x[myxn++]=a[i];
	long long ans=solve(i+1);
	myxn--;
	y[myyn++]=a[i];
	ans+=solve(i+1);
	myyn--;
	return ans;
}

int main()
{
	scanf("%d%s",&n,a);
	myxn=myyn=0;
	printf("%lld\n",solve(0));
	return 0;
}
### Prompt

Your challenge is to write a CPP solution to 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>
using namespace std;
const int N=233,M=20;
char a[N],x[M],y[M];
int myxn,myyn,n;
long long dp[M][M];

long long solve(int i)
{
	if(i==n)
	{
		//printf("233\n");
		memset(dp,0,sizeof(dp));
		dp[0][myxn]=1;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<=myxn;j++)if(dp[i][j])
			{
				if(j && a[i+n]==x[j-1])dp[i+1][j-1]+=dp[i][j];
				int k=myxn-j-i+myyn;
				//printf("i=%d j=%d k=%d\n",i,j,k);
				if(k && a[i+n]==y[k-1])dp[i+1][j]+=dp[i][j];
			}
		}
		return dp[n][0];
	}
	x[myxn++]=a[i];
	long long ans=solve(i+1);
	myxn--;
	y[myyn++]=a[i];
	ans+=solve(i+1);
	myyn--;
	return ans;
}

int main()
{
	scanf("%d%s",&n,a);
	myxn=myyn=0;
	printf("%lld\n",solve(0));
	return 0;
}
```