#include<cstdio>
#include<string>
#include<map>
#include<iostream>
#define maxn 40
#define LL long long
using namespace std;
typedef pair<string,string> PSS;
int n;
LL ans;
string s;
map<PSS,LL> cnt;
void dfs0(string s1,string s2,int i)
{
	if(i==n)
	{
		cnt[PSS(s1,s2)]++;
		return;
	}
	dfs0(s1+s[i],s2,i+1);
	dfs0(s1,s2+s[i],i+1);
}
void dfs1(string s1,string s2,int i)
{
	if(i==n-1)
	{
		if(cnt.count(PSS(s1,s2))) ans+=cnt[PSS(s1,s2)];
		return;
	}
	dfs1(s1+s[i],s2,i-1);
	dfs1(s1,s2+s[i],i-1);
}
int main()
{
	scanf("%d",&n);
	cin>>s;
	string s1,s2;
	dfs0(s1,s2,0);
	dfs1(s1,s2,2*n-1);
	printf("%lld\n",ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<string>
#include<map>
#include<iostream>
#define maxn 40
#define LL long long
using namespace std;
typedef pair<string,string> PSS;
int n;
LL ans;
string s;
map<PSS,LL> cnt;
void dfs0(string s1,string s2,int i)
{
	if(i==n)
	{
		cnt[PSS(s1,s2)]++;
		return;
	}
	dfs0(s1+s[i],s2,i+1);
	dfs0(s1,s2+s[i],i+1);
}
void dfs1(string s1,string s2,int i)
{
	if(i==n-1)
	{
		if(cnt.count(PSS(s1,s2))) ans+=cnt[PSS(s1,s2)];
		return;
	}
	dfs1(s1+s[i],s2,i-1);
	dfs1(s1,s2+s[i],i-1);
}
int main()
{
	scanf("%d",&n);
	cin>>s;
	string s1,s2;
	dfs0(s1,s2,0);
	dfs1(s1,s2,2*n-1);
	printf("%lld\n",ans);
}
```