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

typedef long long ll;
typedef pair <string, string> pss;

const int MAXN = 20;	

int N;
char s[2 * MAXN];
map <pss, int> lft, rig;

int main() {
	scanf("%d%s", &N, s);
	
	for (int i = 0; i < 1 << N; i++) {
		string l1 = "", l2 = "", r1 = "", r2 = "";
		for (int j = 0; j < N; j++) 
			if (i >> j & 1) 
				l1 += s[j];
		for (int j = N - 1; j >= 0; j--)
			if (!(i >> j & 1)) {
				l2 += s[j];
				r1 += s[N + j];
			}
		for (int j = 0; j < N; j++)
			if (i >> j & 1)
				r2 += s[N + j];
		lft[{l1, l2}]++;
		rig[{r1, r2}]++;
	}		
	
	ll ans = 0;
	for (auto it : lft)
		ans += (ll)it.second * rig[it.first];
	
	printf("%lld\n", ans);
	
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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;
typedef pair <string, string> pss;

const int MAXN = 20;	

int N;
char s[2 * MAXN];
map <pss, int> lft, rig;

int main() {
	scanf("%d%s", &N, s);
	
	for (int i = 0; i < 1 << N; i++) {
		string l1 = "", l2 = "", r1 = "", r2 = "";
		for (int j = 0; j < N; j++) 
			if (i >> j & 1) 
				l1 += s[j];
		for (int j = N - 1; j >= 0; j--)
			if (!(i >> j & 1)) {
				l2 += s[j];
				r1 += s[N + j];
			}
		for (int j = 0; j < N; j++)
			if (i >> j & 1)
				r2 += s[N + j];
		lft[{l1, l2}]++;
		rig[{r1, r2}]++;
	}		
	
	ll ans = 0;
	for (auto it : lft)
		ans += (ll)it.second * rig[it.first];
	
	printf("%lld\n", ans);
	
	return 0;
}
```