#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
map <pair<string,string>,int> mp;
int main()
{
	int n;
	long long ans=0;
	string s,a,b;
	cin>>n;
	cin>>s;
	mp.clear();
	for(int i=0;i<(1<<n);i++)
	{
		a="";
		b="";
		for(int j=0;j<n;j++)
			if(i&(1<<j))
				a+=s[j];
			else
				b+=s[j];
		mp[make_pair(a,b)]++;
	}
	for(int i=0;i<(1<<n);i++)
	{
		a="";
		b="";
		for(int j=1;j<=n;j++)
			if(i&(1<<j))
				a+=s[n+n-j];
			else
				b+=s[n+n-j];
		ans+=mp[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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
map <pair<string,string>,int> mp;
int main()
{
	int n;
	long long ans=0;
	string s,a,b;
	cin>>n;
	cin>>s;
	mp.clear();
	for(int i=0;i<(1<<n);i++)
	{
		a="";
		b="";
		for(int j=0;j<n;j++)
			if(i&(1<<j))
				a+=s[j];
			else
				b+=s[j];
		mp[make_pair(a,b)]++;
	}
	for(int i=0;i<(1<<n);i++)
	{
		a="";
		b="";
		for(int j=1;j<=n;j++)
			if(i&(1<<j))
				a+=s[n+n-j];
			else
				b+=s[n+n-j];
		ans+=mp[make_pair(a,b)];
	}
	printf("%lld",ans);
	return 0;
}
```