#include<iostream>
#include<string>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<map>
using namespace std;
map<pair <string,string>,int> f;
int main()
{
	int n;
	scanf("%d",&n);
	string st;
	cin >> st;
	int i,j;
	for (i=0;i<(1<<n);i++)
	{
		string s1="",s2="";
		for (j=0;j<n;j++)
		    if (i&(1<<j)) s1+=st[j];
		    else s2+=st[j];
		f[make_pair(s1,s2)]++;
	}
	
	long long ans=0;
	for (i=0;i<(1<<n);i++)
	{
		string s1="",s2="";
		for (j=n-1;j>=0;j--)
		    if (i&(1<<j)) s2+=st[j+n];
		    else s1+=st[j+n];
		ans+=f[make_pair(s1,s2)];
	}
	printf("%lld",ans);
	return 0;
}
### 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<iostream>
#include<string>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<map>
using namespace std;
map<pair <string,string>,int> f;
int main()
{
	int n;
	scanf("%d",&n);
	string st;
	cin >> st;
	int i,j;
	for (i=0;i<(1<<n);i++)
	{
		string s1="",s2="";
		for (j=0;j<n;j++)
		    if (i&(1<<j)) s1+=st[j];
		    else s2+=st[j];
		f[make_pair(s1,s2)]++;
	}
	
	long long ans=0;
	for (i=0;i<(1<<n);i++)
	{
		string s1="",s2="";
		for (j=n-1;j>=0;j--)
		    if (i&(1<<j)) s2+=st[j+n];
		    else s1+=st[j+n];
		ans+=f[make_pair(s1,s2)];
	}
	printf("%lld",ans);
	return 0;
}
```