#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

map <pair<string, string>, ll> M;

ll n, ans;
char str[44];

int main()
{
	ll i, j;
	
	scanf("%lld%s", &n, str);
	
	for(i=0;i<(1<<n);i++){
		string s1, s2;
		for(j=0;j<n;j++){
			if(i & (1<<j)) s1.push_back(str[j]);
			else s2.push_back(str[j]);
		}
		M[make_pair(s1, s2)] ++;
	}
	
	for(i=0;i<(1<<n);i++){
		string s1, s2;
		for(j=0;j<n;j++){
			if(i & (1<<j)) s1.push_back(str[n+n-1-j]);
			else s2.push_back(str[n+n-1-j]);
		}
		ans += M[make_pair(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;

typedef long long ll;

map <pair<string, string>, ll> M;

ll n, ans;
char str[44];

int main()
{
	ll i, j;
	
	scanf("%lld%s", &n, str);
	
	for(i=0;i<(1<<n);i++){
		string s1, s2;
		for(j=0;j<n;j++){
			if(i & (1<<j)) s1.push_back(str[j]);
			else s2.push_back(str[j]);
		}
		M[make_pair(s1, s2)] ++;
	}
	
	for(i=0;i<(1<<n);i++){
		string s1, s2;
		for(j=0;j<n;j++){
			if(i & (1<<j)) s1.push_back(str[n+n-1-j]);
			else s2.push_back(str[n+n-1-j]);
		}
		ans += M[make_pair(s1, s2)];
	}
	
	printf("%lld\n", ans);
	
	return 0;
}
```