#include <bits/stdc++.h>
using namespace std;
int i,j,k,n,m,x,y,t;
#define mp make_pair
map<pair<string,string>,int>Map;
string s,s1,s2;
int b[19];
long long ans;
int main(){
	scanf("%d\n",&n);
	cin>>s;
	for (int P=0;P<1<<n;P++){
		s1="";s2="";
		x=P;
		for (i=n;i>=1;i--)b[i]=x&1,x>>=1;
		for (i=1;i<=n;i++)
			if (b[i]){
				s1+=s[i+n-1];
			}
			else s2+=s[i+n-1];
		Map[mp(s1,s2)]++;
	}
	for (int P=0;P<1<<n;P++){
		x=P;
		for (i=n;i>=1;i--)b[i]=x&1,x>>=1;
		s1="";s2="";
		for (i=1;i<=n;i++)if (b[i])s1=s[i-1]+s1;else s2=s[i-1]+s2;
		ans+=Map[mp(s1,s2)];
	}
	printf("%lld\n",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 <bits/stdc++.h>
using namespace std;
int i,j,k,n,m,x,y,t;
#define mp make_pair
map<pair<string,string>,int>Map;
string s,s1,s2;
int b[19];
long long ans;
int main(){
	scanf("%d\n",&n);
	cin>>s;
	for (int P=0;P<1<<n;P++){
		s1="";s2="";
		x=P;
		for (i=n;i>=1;i--)b[i]=x&1,x>>=1;
		for (i=1;i<=n;i++)
			if (b[i]){
				s1+=s[i+n-1];
			}
			else s2+=s[i+n-1];
		Map[mp(s1,s2)]++;
	}
	for (int P=0;P<1<<n;P++){
		x=P;
		for (i=n;i>=1;i--)b[i]=x&1,x>>=1;
		s1="";s2="";
		for (i=1;i<=n;i++)if (b[i])s1=s[i-1]+s1;else s2=s[i-1]+s2;
		ans+=Map[mp(s1,s2)];
	}
	printf("%lld\n",ans);
	return 0;
}

```