#include <bits/stdc++.h>
using namespace std;
int main(){
	int n; cin >> n;
	string s; cin >> s;
	map<string, int> ways[20];
	for(int i = 0; i < 1<<n; i++){
		string in0, in1;
		for(int j = 0; j < n; j++){
			if(i&(1<<j)) in1.push_back(s[j]);
			else in0.push_back(s[j]);
		}
		reverse(in1.begin(), in1.end());
		ways[in0.size()][in0 + in1]++;
	}
	long long ans = 0;
	for(int i = 0; i < 1<<n; i++){
		string in0, in1;
		for(int j = 0; j < n; j++){
			if(i&(1<<j)) in1.push_back(s[j + n]);
			else in0.push_back(s[j + n]);
		}
		reverse(in1.begin(), in1.end());
		ans += ways[n - in0.size()][in1 + in0];
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Please provide a cpp coded solution 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 <bits/stdc++.h>
using namespace std;
int main(){
	int n; cin >> n;
	string s; cin >> s;
	map<string, int> ways[20];
	for(int i = 0; i < 1<<n; i++){
		string in0, in1;
		for(int j = 0; j < n; j++){
			if(i&(1<<j)) in1.push_back(s[j]);
			else in0.push_back(s[j]);
		}
		reverse(in1.begin(), in1.end());
		ways[in0.size()][in0 + in1]++;
	}
	long long ans = 0;
	for(int i = 0; i < 1<<n; i++){
		string in0, in1;
		for(int j = 0; j < n; j++){
			if(i&(1<<j)) in1.push_back(s[j + n]);
			else in0.push_back(s[j + n]);
		}
		reverse(in1.begin(), in1.end());
		ans += ways[n - in0.size()][in1 + in0];
	}
	cout << ans << endl;
	return 0;
}

```