#include<cstdio>
#include<vector>
#include<cstring>
#include<string>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
long long n,ans;
char s[50];
map<pair<string,string>,long long>M;
void dfs1(string s1,string s2,long long x)
{
	if(x==n)
	{
		pair<string,string> o;
		o.first=s1;
		o.second=s2;
		M[o]++;
		return ;
	}
	dfs1(s1+s[x],s2,x+1);
	dfs1(s1,s2+s[x],x+1);
}
void dfs2(string s1,string s2,long long x)
{
	if(x==n-1)
	{
		pair<string,string> o;
		o.first=s1;
		o.second=s2;
		ans+=M[o];
		return ;
	}
	dfs2(s1+s[x],s2,x-1);
	dfs2(s1,s2+s[x],x-1);
}
int main()
{
	scanf("%lld",&n);
	scanf("%s",s);
	string s1,s2;
	dfs1(s1,s2,0);
	dfs2(s1,s2,2*n-1);
	printf("%lld\n",ans);
}
### 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<cstdio>
#include<vector>
#include<cstring>
#include<string>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
long long n,ans;
char s[50];
map<pair<string,string>,long long>M;
void dfs1(string s1,string s2,long long x)
{
	if(x==n)
	{
		pair<string,string> o;
		o.first=s1;
		o.second=s2;
		M[o]++;
		return ;
	}
	dfs1(s1+s[x],s2,x+1);
	dfs1(s1,s2+s[x],x+1);
}
void dfs2(string s1,string s2,long long x)
{
	if(x==n-1)
	{
		pair<string,string> o;
		o.first=s1;
		o.second=s2;
		ans+=M[o];
		return ;
	}
	dfs2(s1+s[x],s2,x-1);
	dfs2(s1,s2+s[x],x-1);
}
int main()
{
	scanf("%lld",&n);
	scanf("%s",s);
	string s1,s2;
	dfs1(s1,s2,0);
	dfs2(s1,s2,2*n-1);
	printf("%lld\n",ans);
}
```