#include <bits/stdc++.h>

#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 38;

char S[N];
int n;
map<string, int> M[N];

int main() {

	scanf("%d%s", &n, S);
	For(i, 0, (1 << n) - 1) {
		char T[20];
		int l = 0, r = n;
		For(j, 0, n - 1) if (i & (1 << j)) T[l++] = S[j]; else T[--r] = S[j];
		T[n] = '\0';
		M[l][string(T)]++;
	}

	long long ans = 0;
	For(i, 0, (1 << n) - 1) {
		char T[20];
		int l = 0, r = n;
		For(j, 0, n - 1) if (i & (1 << j)) T[l++] = S[2 * n - 1 - j]; else T[--r] = S[2 * n - 1 - j];
		T[n] = '\0';
		ans += M[l][string(T)];
	}
	printf("%lld\n", ans);
	
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>

#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 38;

char S[N];
int n;
map<string, int> M[N];

int main() {

	scanf("%d%s", &n, S);
	For(i, 0, (1 << n) - 1) {
		char T[20];
		int l = 0, r = n;
		For(j, 0, n - 1) if (i & (1 << j)) T[l++] = S[j]; else T[--r] = S[j];
		T[n] = '\0';
		M[l][string(T)]++;
	}

	long long ans = 0;
	For(i, 0, (1 << n) - 1) {
		char T[20];
		int l = 0, r = n;
		For(j, 0, n - 1) if (i & (1 << j)) T[l++] = S[2 * n - 1 - j]; else T[--r] = S[2 * n - 1 - j];
		T[n] = '\0';
		ans += M[l][string(T)];
	}
	printf("%lld\n", ans);
	
	return 0;
}
```