// by piano
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=30;
int n,op[N];
char s[N];
map <string,int> tot;
int main(){
	scanf("%d",&n);
	scanf("%s",s+1);
	tot.clear();
	for (int i=0;i<(1<<n);i++){
		string now="";
		for (int j=0;j<n;j++)
			if (!((i>>j)&1))
				now+=s[j+1];
		now+="*";
		for (int j=n-1;j>=0;j--)
			if ((i>>j)&1)
				now+=s[j+1];
//		cout << now << endl;
		tot[now]++;
	}
//	cout << endl;
	LL ans=0;
	for (int i=0;i<(1<<n);i++){
		string now="";
		for (int j=0;j<n;j++)
			if ((i>>j)&1)
				now+=s[n*2-j];
		now+="*";
		for (int j=n-1;j>=0;j--)
			if (!((i>>j)&1))
				now+=s[n*2-j];
//		cout << now << tot[now] << endl;
		ans+=tot[now];
	}
	printf("%lld",ans);
	return 0;
}
### Prompt

Your task is to create a CPP solution to 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
// by piano
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=30;
int n,op[N];
char s[N];
map <string,int> tot;
int main(){
	scanf("%d",&n);
	scanf("%s",s+1);
	tot.clear();
	for (int i=0;i<(1<<n);i++){
		string now="";
		for (int j=0;j<n;j++)
			if (!((i>>j)&1))
				now+=s[j+1];
		now+="*";
		for (int j=n-1;j>=0;j--)
			if ((i>>j)&1)
				now+=s[j+1];
//		cout << now << endl;
		tot[now]++;
	}
//	cout << endl;
	LL ans=0;
	for (int i=0;i<(1<<n);i++){
		string now="";
		for (int j=0;j<n;j++)
			if ((i>>j)&1)
				now+=s[n*2-j];
		now+="*";
		for (int j=n-1;j>=0;j--)
			if (!((i>>j)&1))
				now+=s[n*2-j];
//		cout << now << tot[now] << endl;
		ans+=tot[now];
	}
	printf("%lld",ans);
	return 0;
}
```