#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#define ll long long
#define F(i,s,t) for(int i=s;i<=t;i++)
#define DF(i,s,t) for(int i=s;i>=t;i--)
#define mp make_pair
using namespace std;
map<pair<string,string>,ll>sa;
char s[233];
ll sum=0;
int n;
string a,b;

int main(){
	scanf("%d",&n);
    scanf("%s",s);
    F(i,0,(1<<n)-1){
        a="",b="";
        F(j,0,n-1) (i&(1<<j))?(a+=s[j]):(b+=s[j]);
        sa[mp(a,b)]++;
    }
    F(i,0,(1<<n)-1){
        a="",b="";
        DF(j,n-1,0) (i&(1<<j))?(b+=s[j+n]):(a+=s[j+n]);
        sum+=sa[mp(a,b)];
    }
    printf("%lld\n",sum);
    return 0;
}

/*


*/

### Prompt

Generate 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
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#define ll long long
#define F(i,s,t) for(int i=s;i<=t;i++)
#define DF(i,s,t) for(int i=s;i>=t;i--)
#define mp make_pair
using namespace std;
map<pair<string,string>,ll>sa;
char s[233];
ll sum=0;
int n;
string a,b;

int main(){
	scanf("%d",&n);
    scanf("%s",s);
    F(i,0,(1<<n)-1){
        a="",b="";
        F(j,0,n-1) (i&(1<<j))?(a+=s[j]):(b+=s[j]);
        sa[mp(a,b)]++;
    }
    F(i,0,(1<<n)-1){
        a="",b="";
        DF(j,n-1,0) (i&(1<<j))?(b+=s[j+n]):(a+=s[j+n]);
        sum+=sa[mp(a,b)];
    }
    printf("%lld\n",sum);
    return 0;
}

/*


*/

```