#include<map>
#include<cstdio>
#include<algorithm>
#include<string>
using namespace std;
int n;
long long ans;
char str[10005];
map<pair<string,string>,int>f;

int main()
{
	scanf("%d%s",&n,str);
	int mxs = 1<<n;
	for(int s = 0; s < mxs; s++)
	{
		string a,b;
		for(int j = 0; j < n; j++)
		{
			if((s>>j)&1) a += str[j];
			else b += str[j];
		}
		f[make_pair(a,b)]++;
	}
	for(int s = 0; s < mxs; s++)
	{
		string a,b;
		for(int j = 0; j < n; j++)
		{
			if((s>>j)&1) a += str[(n<<1)-1-j];
			else b += str[(n<<1)-1-j];
		}
		ans+=f[make_pair(a,b)];
	}
	printf("%lld",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<map>
#include<cstdio>
#include<algorithm>
#include<string>
using namespace std;
int n;
long long ans;
char str[10005];
map<pair<string,string>,int>f;

int main()
{
	scanf("%d%s",&n,str);
	int mxs = 1<<n;
	for(int s = 0; s < mxs; s++)
	{
		string a,b;
		for(int j = 0; j < n; j++)
		{
			if((s>>j)&1) a += str[j];
			else b += str[j];
		}
		f[make_pair(a,b)]++;
	}
	for(int s = 0; s < mxs; s++)
	{
		string a,b;
		for(int j = 0; j < n; j++)
		{
			if((s>>j)&1) a += str[(n<<1)-1-j];
			else b += str[(n<<1)-1-j];
		}
		ans+=f[make_pair(a,b)];
	}
	printf("%lld",ans);
	return 0;
}
```