#include <cstdio>
#include <string>
#include <map>

using namespace std;

char s[37];

int main() {
	int n, i, j;
	long long ans = 0;
	map <pair<string, string>, long long> mp;
	
	scanf("%d", &n);
	scanf("%s", s);
	
	for (i = 0; i < (1 << n); i++) {
		string s1 = "", s2 = "";
		
		for (j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				s1 += s[j];
			} else {
				s2 += s[j];
			}
		}
		
		mp[make_pair(s1, s2)]++;
	}
	
	for (i = 0; i < (1 << n); i++) {
		string s1 = "", s2 = "";
		
		for (j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				s1 += s[n * 2 - j - 1];
			} else {
				s2 += s[n * 2 - j - 1];
			}
		}
		
		ans += mp[make_pair(s2, s1)];
	}
	
	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 <cstdio>
#include <string>
#include <map>

using namespace std;

char s[37];

int main() {
	int n, i, j;
	long long ans = 0;
	map <pair<string, string>, long long> mp;
	
	scanf("%d", &n);
	scanf("%s", s);
	
	for (i = 0; i < (1 << n); i++) {
		string s1 = "", s2 = "";
		
		for (j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				s1 += s[j];
			} else {
				s2 += s[j];
			}
		}
		
		mp[make_pair(s1, s2)]++;
	}
	
	for (i = 0; i < (1 << n); i++) {
		string s1 = "", s2 = "";
		
		for (j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				s1 += s[n * 2 - j - 1];
			} else {
				s2 += s[n * 2 - j - 1];
			}
		}
		
		ans += mp[make_pair(s2, s1)];
	}
	
	printf("%lld\n", ans);
	
	return 0;
}

```