#include <math.h>
#include <bits/stdc++.h>//一括で読み込み
typedef long long ll;
using namespace std;

map<pair<string, string>,int> g;
int n;
string s;
ll answer;
int main()
{
	cin >> n >> s;
  
	for(int i = 0; i < (1<<n); i++)//文字数の半分をbit
  {
		string a,b;
		for(int j = 0; j < n; j++)
    {
			if((i>>j) & 1) a += s[j];
			else b += s[j];
		}
		g[{a, b}]++;
	}
  
	for(int i = 0; i < (1<<n); i++)//文字数の半分をbit
  {
		string a,b;
		for(int j = 0; j < n; j++)
    {
			if((i>>j) & 1) a += s[2*n-1-j];//逆から数えてj文字目
			else b += s[2*n-1-j];
		}
		answer += g[{a,b}];
	}
  
	cout << answer << endl;
	return 0;
}

### Prompt

Please create a solution in cpp to 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 <math.h>
#include <bits/stdc++.h>//一括で読み込み
typedef long long ll;
using namespace std;

map<pair<string, string>,int> g;
int n;
string s;
ll answer;
int main()
{
	cin >> n >> s;
  
	for(int i = 0; i < (1<<n); i++)//文字数の半分をbit
  {
		string a,b;
		for(int j = 0; j < n; j++)
    {
			if((i>>j) & 1) a += s[j];
			else b += s[j];
		}
		g[{a, b}]++;
	}
  
	for(int i = 0; i < (1<<n); i++)//文字数の半分をbit
  {
		string a,b;
		for(int j = 0; j < n; j++)
    {
			if((i>>j) & 1) a += s[2*n-1-j];//逆から数えてj文字目
			else b += s[2*n-1-j];
		}
		answer += g[{a,b}];
	}
  
	cout << answer << endl;
	return 0;
}

```