#include <bits/stdc++.h>

using namespace std;

int n;

char str[50];

map<pair<string, string>, int> ss; 

int main()
{
	scanf("%d", &n);
	scanf("%s", str);
	for (int i = 0; i < (1 << n); ++i)
	{
		string s, t;
		s.clear(), t.clear();
		for (int j = 0; j < n; ++j)
			if (i & (1 << j))
				s.push_back(str[j]);
			else
				t.push_back(str[j]);
		++ss[make_pair(s, t)];
	}
	long long ans = 0;
	for (int i = 0; i < (1 << n); ++i)
	{
		string s, t;
		s.clear(), t.clear();
		for (int j = 0; j < n; ++j)
			if (i & (1 << j))
				s.push_back(str[(n << 1) - 1 - j]);
			else
				t.push_back(str[(n << 1) - 1 - j]);
		ans += ss[make_pair(s, t)];
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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;

int n;

char str[50];

map<pair<string, string>, int> ss; 

int main()
{
	scanf("%d", &n);
	scanf("%s", str);
	for (int i = 0; i < (1 << n); ++i)
	{
		string s, t;
		s.clear(), t.clear();
		for (int j = 0; j < n; ++j)
			if (i & (1 << j))
				s.push_back(str[j]);
			else
				t.push_back(str[j]);
		++ss[make_pair(s, t)];
	}
	long long ans = 0;
	for (int i = 0; i < (1 << n); ++i)
	{
		string s, t;
		s.clear(), t.clear();
		for (int j = 0; j < n; ++j)
			if (i & (1 << j))
				s.push_back(str[(n << 1) - 1 - j]);
			else
				t.push_back(str[(n << 1) - 1 - j]);
		ans += ss[make_pair(s, t)];
	}
	printf("%lld\n", ans);
	return 0;
}
```