#include <bits/stdc++.h>

using namespace std;

map<pair<string, string>, long long> cnt1, cnt2;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int n, i;
	string s;
	string a, b, c, d;
	long long sol = 0;
	
	cin>>n>>s;
	
	for (int bit = 0; bit < (1<<n); ++bit)
	{
		a = b = c = d = "";
		
		for (i = 0; i < n; ++i)
		{
			(bit & 1<<i ? a : b) += s[i];
			(bit & 1<<i ? d : c) += s[2 * n - i - 1];
		}
		
		++cnt1[{a, b}];
		++cnt2[{c, d}];
	}
	
	for (const auto& x: cnt1)
	{
//		cout<<x.first.first<<' '<<x.first.second<<' '<<x.second<<' '<<cnt2[x.first]<<endl;
		sol += x.second * cnt2[x.first];
	}
	
	cout<<sol<<'\n';
}

### 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;

map<pair<string, string>, long long> cnt1, cnt2;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int n, i;
	string s;
	string a, b, c, d;
	long long sol = 0;
	
	cin>>n>>s;
	
	for (int bit = 0; bit < (1<<n); ++bit)
	{
		a = b = c = d = "";
		
		for (i = 0; i < n; ++i)
		{
			(bit & 1<<i ? a : b) += s[i];
			(bit & 1<<i ? d : c) += s[2 * n - i - 1];
		}
		
		++cnt1[{a, b}];
		++cnt2[{c, d}];
	}
	
	for (const auto& x: cnt1)
	{
//		cout<<x.first.first<<' '<<x.first.second<<' '<<x.second<<' '<<cnt2[x.first]<<endl;
		sol += x.second * cnt2[x.first];
	}
	
	cout<<sol<<'\n';
}

```