#include <bits/stdc++.h>

using namespace std;

map<string, int> mp;

int main()
{
	int n;
	cin >> n;
	string s;
	cin >> s;
	for(int bin=0; bin<(1<<n); bin++)
	{
		string left="", right="";
		for(int i=0; i<n; i++)
		{
			if((bin & (1<<i))>>i) left=left+s[i];
			else right=s[i]+right;
		}
		string t=left+"0"+right;
		mp[t]++;
	}
	long long ans=0;
	for(int bin=0; bin<(1<<n); bin++)
	{
		string left="", right="";
		for(int i=0; i<n; i++)
		{
			if((bin & (1<<i))>>i) left=left+s[2*n-1-i];
			else right=s[2*n-1-i]+right;
		}
		string t=left+"0"+right;
		ans+=(long long)mp[t];
	}
	cout << ans << endl;
	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 <bits/stdc++.h>

using namespace std;

map<string, int> mp;

int main()
{
	int n;
	cin >> n;
	string s;
	cin >> s;
	for(int bin=0; bin<(1<<n); bin++)
	{
		string left="", right="";
		for(int i=0; i<n; i++)
		{
			if((bin & (1<<i))>>i) left=left+s[i];
			else right=s[i]+right;
		}
		string t=left+"0"+right;
		mp[t]++;
	}
	long long ans=0;
	for(int bin=0; bin<(1<<n); bin++)
	{
		string left="", right="";
		for(int i=0; i<n; i++)
		{
			if((bin & (1<<i))>>i) left=left+s[2*n-1-i];
			else right=s[2*n-1-i]+right;
		}
		string t=left+"0"+right;
		ans+=(long long)mp[t];
	}
	cout << ans << endl;
	return 0;
}
```