#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int N;
char S[40], curr[20];
ll dp[20][20], tot;

int main() {
	scanf("%d", &N);
	scanf("%s", S);

	for (int msk = 0; msk < (1<<N); ++msk) {
		int cnt = 0, bcnt = 0;
		for (int i = 0; i < N; ++i) if ((msk>>i)&1) curr[cnt++] = S[i], bcnt++;
		for (int i = N; i--; ) if (!((msk>>i)&1)) curr[cnt++] = S[i];

		for (int i = 0; i < 20; ++i) for (int j = 0; j < 20; ++j) dp[i][j] = 0;

		dp[bcnt][N-bcnt] = 1;
		for (int i = 0; i <= N; ++i) for (int j = 0; j <= N; ++j) {
			if (i < N && S[i+j] == curr[i]) dp[i+1][j] += dp[i][j];
			if (j < N && S[i+j] == curr[N-j-1]) dp[i][j+1] += dp[i][j];
		}

		tot += dp[N][N];
	}

	printf("%lld\n", tot);
}
### Prompt

Create a solution in CPP for 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 N;
char S[40], curr[20];
ll dp[20][20], tot;

int main() {
	scanf("%d", &N);
	scanf("%s", S);

	for (int msk = 0; msk < (1<<N); ++msk) {
		int cnt = 0, bcnt = 0;
		for (int i = 0; i < N; ++i) if ((msk>>i)&1) curr[cnt++] = S[i], bcnt++;
		for (int i = N; i--; ) if (!((msk>>i)&1)) curr[cnt++] = S[i];

		for (int i = 0; i < 20; ++i) for (int j = 0; j < 20; ++j) dp[i][j] = 0;

		dp[bcnt][N-bcnt] = 1;
		for (int i = 0; i <= N; ++i) for (int j = 0; j <= N; ++j) {
			if (i < N && S[i+j] == curr[i]) dp[i+1][j] += dp[i][j];
			if (j < N && S[i+j] == curr[N-j-1]) dp[i][j+1] += dp[i][j];
		}

		tot += dp[N][N];
	}

	printf("%lld\n", tot);
}
```