#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
typedef long long LL;
int n;
LL ans;
char s[40];
map<pair<string, string>, LL>cnt;
int main() {
	scanf("%d", &n);
	scanf("%s", s);
	string s1, s2;
	for(int i = 0; i < (1 << n); i++) {
		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)]++;
	}
	for(int i = 0; i < (1 << n); i++) {
		s1 = s2 = "";
		for(int j = 0; j < n; j++) {
			if(i & (1 << j))s1 += s[2 * n - j - 1];
			else s2 += s[2 * n - j - 1];
		}
		ans += cnt[make_pair(s1, s2)];
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
typedef long long LL;
int n;
LL ans;
char s[40];
map<pair<string, string>, LL>cnt;
int main() {
	scanf("%d", &n);
	scanf("%s", s);
	string s1, s2;
	for(int i = 0; i < (1 << n); i++) {
		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)]++;
	}
	for(int i = 0; i < (1 << n); i++) {
		s1 = s2 = "";
		for(int j = 0; j < n; j++) {
			if(i & (1 << j))s1 += s[2 * n - j - 1];
			else s2 += s[2 * n - j - 1];
		}
		ans += cnt[make_pair(s1, s2)];
	}
	printf("%lld\n", ans);
	return 0;
}
```