#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <iostream>

using namespace std;
typedef long long LL;

map<pair<string, string>, LL> cnt;
const int maxn = 50;
char s[maxn];

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

	for (int i = 0; i < (1 << n); ++i) {
		string s1 = "", s2 = "";
		for (int j = 0; j < n; ++j) {
			if (i & (1 << j)) s1 += s[j];
			else s2 += s[j];
		}
		cnt[make_pair(s1, s2)]++;
	}

	LL sum = 0;

	for (int i = 0; i < (1 << n); ++i) {
		string s1 = "", s2 = "";

		for (int j = n - 1; j >= 0; --j) {
			if (i & (1 << j)) s2 += s[j + n];
			else s1 += s[j + n];
		}

		sum += cnt[make_pair(s1, s2)];
	}

	printf("%lld\n", sum);

	return 0;
}
### Prompt

Your task is to create 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 <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <iostream>

using namespace std;
typedef long long LL;

map<pair<string, string>, LL> cnt;
const int maxn = 50;
char s[maxn];

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

	for (int i = 0; i < (1 << n); ++i) {
		string s1 = "", s2 = "";
		for (int j = 0; j < n; ++j) {
			if (i & (1 << j)) s1 += s[j];
			else s2 += s[j];
		}
		cnt[make_pair(s1, s2)]++;
	}

	LL sum = 0;

	for (int i = 0; i < (1 << n); ++i) {
		string s1 = "", s2 = "";

		for (int j = n - 1; j >= 0; --j) {
			if (i & (1 << j)) s2 += s[j + n];
			else s1 += s[j + n];
		}

		sum += cnt[make_pair(s1, s2)];
	}

	printf("%lld\n", sum);

	return 0;
}
```