#include <bits/stdc++.h>

using namespace std;

const int MX = 18;

char s[2 * MX + 1], a[MX], b[MX];
long long dp[MX][MX];

int main() {
	int n;
	ignore = scanf("%d %s", &n, s);
	
	long long ans = 0;
	for (int msk = 0; msk < (1 << n); msk++) {
		int p = 0, q = 0;
		for (int i = 2 * n - 1, m = msk; i >= n; i--, m /= 2) {
			if (m & 1) a[p++] = s[i];
			else b[q++] = s[i];
		}
		
		dp[0][0] = 1;
		for (int i = 0; i <= p; i++)
			for (int j = 0; j <= q; j++) {
				if (i + j == 0) continue;
				dp[i][j] = 0;
				if (i > 0 && s[i + j - 1] == a[i - 1]) dp[i][j] += dp[i - 1][j];
				if (j > 0 && s[i + j - 1] == b[j - 1]) dp[i][j] += dp[i][j - 1];
			}
		
		ans += dp[p][q];
	}
	
	printf("%lld\n", ans);
	
	return 0;
}

### Prompt

Please formulate 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;

const int MX = 18;

char s[2 * MX + 1], a[MX], b[MX];
long long dp[MX][MX];

int main() {
	int n;
	ignore = scanf("%d %s", &n, s);
	
	long long ans = 0;
	for (int msk = 0; msk < (1 << n); msk++) {
		int p = 0, q = 0;
		for (int i = 2 * n - 1, m = msk; i >= n; i--, m /= 2) {
			if (m & 1) a[p++] = s[i];
			else b[q++] = s[i];
		}
		
		dp[0][0] = 1;
		for (int i = 0; i <= p; i++)
			for (int j = 0; j <= q; j++) {
				if (i + j == 0) continue;
				dp[i][j] = 0;
				if (i > 0 && s[i + j - 1] == a[i - 1]) dp[i][j] += dp[i - 1][j];
				if (j > 0 && s[i + j - 1] == b[j - 1]) dp[i][j] += dp[i][j - 1];
			}
		
		ans += dp[p][q];
	}
	
	printf("%lld\n", ans);
	
	return 0;
}

```