#include <iostream>
#include <string>
#include <map>
#include <algorithm>

using namespace std;

typedef long long ll;

string S, Sr;
int N;
map<string, ll> mp;
string red, blue;
ll ans;

void dfs(ll a) {
	if(a == N) {
		mp[red + '-' + blue]++;
		return ;
	}
	red += S[a];
	dfs(a + 1);
	red.pop_back();
	blue += S[a];
	dfs(a + 1);
	blue.pop_back();
}

void dfs2(ll a) {
	if(a == N) {
		ans += mp[blue + '-' + red];
		return ;
	}
	red += Sr[a];
	dfs2(a + 1);
	red.pop_back();
	blue += Sr[a];
	dfs2(a + 1);
	blue.pop_back();
}

int main()
{
	cin >> N;
	cin >> S;
	Sr = S;
	reverse(Sr.begin(), Sr.end());
	red = blue = "";
	dfs(0);
	red = blue = "";
	dfs2(0);
	cout << ans << endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 <iostream>
#include <string>
#include <map>
#include <algorithm>

using namespace std;

typedef long long ll;

string S, Sr;
int N;
map<string, ll> mp;
string red, blue;
ll ans;

void dfs(ll a) {
	if(a == N) {
		mp[red + '-' + blue]++;
		return ;
	}
	red += S[a];
	dfs(a + 1);
	red.pop_back();
	blue += S[a];
	dfs(a + 1);
	blue.pop_back();
}

void dfs2(ll a) {
	if(a == N) {
		ans += mp[blue + '-' + red];
		return ;
	}
	red += Sr[a];
	dfs2(a + 1);
	red.pop_back();
	blue += Sr[a];
	dfs2(a + 1);
	blue.pop_back();
}

int main()
{
	cin >> N;
	cin >> S;
	Sr = S;
	reverse(Sr.begin(), Sr.end());
	red = blue = "";
	dfs(0);
	red = blue = "";
	dfs2(0);
	cout << ans << endl;
}

```