#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
using namespace std;
struct node
{
	long long x,y;
}a[1000005],b[1000005];
char s[1000005];
map<pair<long long,long long>,long long>Map;
int n,m;
long long ans;
signed main()
{
	scanf("%d",&n);
	scanf("%s",s);
	m=1ll<<n;
	for (int l=0;l<m;l++)
	{
		for (int i=0;i<n;i++)
			if (l>>i&1)	a[l].x=a[l].x*27+s[i]-96;
			else a[l].y=a[l].y*27+s[i]-96;
		for (int i=n-1;i>=0;i--)
			if (l>>i&1) b[l].x=b[l].x*27+s[i+n]-96;
			else b[l].y=b[l].y*27+s[i+n]-96;
	}
	for (int i=0;i<m;i++)
		Map[make_pair(a[i].x,a[i].y)]++;
	for (int i=0;i<m;i++)
		ans+=Map[make_pair(b[i].x,b[i].y)];
	printf("%lld\n",ans);
}

### Prompt

Please create a solution in cpp 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
using namespace std;
struct node
{
	long long x,y;
}a[1000005],b[1000005];
char s[1000005];
map<pair<long long,long long>,long long>Map;
int n,m;
long long ans;
signed main()
{
	scanf("%d",&n);
	scanf("%s",s);
	m=1ll<<n;
	for (int l=0;l<m;l++)
	{
		for (int i=0;i<n;i++)
			if (l>>i&1)	a[l].x=a[l].x*27+s[i]-96;
			else a[l].y=a[l].y*27+s[i]-96;
		for (int i=n-1;i>=0;i--)
			if (l>>i&1) b[l].x=b[l].x*27+s[i+n]-96;
			else b[l].y=b[l].y*27+s[i+n]-96;
	}
	for (int i=0;i<m;i++)
		Map[make_pair(a[i].x,a[i].y)]++;
	for (int i=0;i<m;i++)
		ans+=Map[make_pair(b[i].x,b[i].y)];
	printf("%lld\n",ans);
}

```