#include "bits/stdc++.h"
using namespace std;
typedef long long ll;

int main(void){
	int n,m,i,j,k,r,b;
	char d[2][20],s[44];
	ll dp[20][20],ans=0ll;
	scanf("%d",&n);
	scanf("%s",s);
	m = (1<<n);
	for(i=0; i<m; ++i){
		r = 0; b = 0;
		for(j=0; j<n; ++j){
			if((i&(1<<j))){
				d[0][r] = s[j];
				++r;
			}else{
				d[1][b] = s[j];
				++b;
			}
		}
		fill(dp[0],dp[20],0ll);
		dp[r][b] = 1ll;
		for(j=r; j>=0; --j){
			for(k=b; k>=0; --k){
				if(j > 0 && s[n+j+k-1] == d[0][r-j]){
					dp[j-1][k] += dp[j][k];
				}
				if(k > 0 && s[n+j+k-1] == d[1][b-k]){
					dp[j][k-1] += dp[j][k];
				}
			}
		}
		ans += dp[0][0];
	}
	printf("%lld\n",ans);
	return 0;
}

### Prompt

Your challenge is to write 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 "bits/stdc++.h"
using namespace std;
typedef long long ll;

int main(void){
	int n,m,i,j,k,r,b;
	char d[2][20],s[44];
	ll dp[20][20],ans=0ll;
	scanf("%d",&n);
	scanf("%s",s);
	m = (1<<n);
	for(i=0; i<m; ++i){
		r = 0; b = 0;
		for(j=0; j<n; ++j){
			if((i&(1<<j))){
				d[0][r] = s[j];
				++r;
			}else{
				d[1][b] = s[j];
				++b;
			}
		}
		fill(dp[0],dp[20],0ll);
		dp[r][b] = 1ll;
		for(j=r; j>=0; --j){
			for(k=b; k>=0; --k){
				if(j > 0 && s[n+j+k-1] == d[0][r-j]){
					dp[j-1][k] += dp[j][k];
				}
				if(k > 0 && s[n+j+k-1] == d[1][b-k]){
					dp[j][k-1] += dp[j][k];
				}
			}
		}
		ans += dp[0][0];
	}
	printf("%lld\n",ans);
	return 0;
}

```