#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pss = pair<string, string>;
int n;
const int maxn = 50;
map<pss, int> M;
char s[maxn];
ll ans;
int main()
{
	scanf("%d", &n);
	scanf("%s", s);
	int S = (1 << n) - 1;
	for(int i = 0; i <= S; ++i)
	{
		pss ret;
		for(int j = 0; j < n; ++j)
			if(i & (1 << j))
				ret.first += s[j];
			else 
				ret.second += s[j];
		++M[ret];
	}
	for(int i = 0; i <= S; ++i)
	{
		pss ret;
		for(int j = 0; j < n; ++j)
			if(i & (1 << j))
				ret.second += s[j + n];
			else
				ret.first += s[j + n];
		reverse(ret.first.begin(), ret.first.end());
		reverse(ret.second.begin(), ret.second.end());
		ans += M[ret];
	}
	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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using pss = pair<string, string>;
int n;
const int maxn = 50;
map<pss, int> M;
char s[maxn];
ll ans;
int main()
{
	scanf("%d", &n);
	scanf("%s", s);
	int S = (1 << n) - 1;
	for(int i = 0; i <= S; ++i)
	{
		pss ret;
		for(int j = 0; j < n; ++j)
			if(i & (1 << j))
				ret.first += s[j];
			else 
				ret.second += s[j];
		++M[ret];
	}
	for(int i = 0; i <= S; ++i)
	{
		pss ret;
		for(int j = 0; j < n; ++j)
			if(i & (1 << j))
				ret.second += s[j + n];
			else
				ret.first += s[j + n];
		reverse(ret.first.begin(), ret.first.end());
		reverse(ret.second.begin(), ret.second.end());
		ans += M[ret];
	}
	printf("%lld\n", ans);
	return 0;
}
```