#include <bits/stdc++.h>

#define PI 3.14159265358979323846264338327950288419

using namespace std;

map<pair<string, string>, int> M1;

int n;
char str[39];

int main(){
	scanf("%d %s", &n, str);
	for(int i = 0; i < (1 << n); i++){
		string a = "";
		string b = "";
		for(int j = 0; j < n; j++){
			if(i & (1 << j)){
				a += str[j];
			}else{
				b += str[j];
			}
		}
		if(!M1.count({a, b})){
			M1[{a, b}] = 1;
		}else{
			M1[{a, b}]++;
		}
	}
	
	long long ans = 0;
	
	for(int i = 0; i < (1 << n); i++){
		string a = "";
		string b = "";
		for(int j = n - 1; j >= 0; j--){
			if(i & (1 << j)){
				a += str[j + n];
			}else{
				b += str[j + n];
			}
		}
		if(M1.count({a, b})){
			ans += M1[{a, b}];
		}
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Your task is to create a CPP solution 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 <bits/stdc++.h>

#define PI 3.14159265358979323846264338327950288419

using namespace std;

map<pair<string, string>, int> M1;

int n;
char str[39];

int main(){
	scanf("%d %s", &n, str);
	for(int i = 0; i < (1 << n); i++){
		string a = "";
		string b = "";
		for(int j = 0; j < n; j++){
			if(i & (1 << j)){
				a += str[j];
			}else{
				b += str[j];
			}
		}
		if(!M1.count({a, b})){
			M1[{a, b}] = 1;
		}else{
			M1[{a, b}]++;
		}
	}
	
	long long ans = 0;
	
	for(int i = 0; i < (1 << n); i++){
		string a = "";
		string b = "";
		for(int j = n - 1; j >= 0; j--){
			if(i & (1 << j)){
				a += str[j + n];
			}else{
				b += str[j + n];
			}
		}
		if(M1.count({a, b})){
			ans += M1[{a, b}];
		}
	}
	printf("%lld\n", ans);
	return 0;
}
```