/*input
18
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
*/
#include <bits/stdc++.h>
using namespace std;
int64_t n,ans;
string s;
map<pair<string,string>,int> l[19];
int main(){
	cin>>n>>s;
	for(int mask=0;mask<(1<<n);++mask){
		string red="",blue="";
		int num = __builtin_popcount(mask);
		for(int i=0;i<n;++i) 
			if((mask>>i)&1)
				red += s[i];
			else
				blue += s[i];
		reverse(blue.begin(),blue.end());
		++l[num][{red,blue}];
	}
	for(int mask=0;mask<(1<<n);++mask){
		string red="",blue="";
		int num = __builtin_popcount(mask);
		for(int i=0;i<n;++i) 
			if((mask>>i)&1)
				blue += s[2*n-1-i];
			else
				red += s[2*n-1-i];
		reverse(red.begin(),red.end());
		ans += l[num][{blue,red}];
	}
	cout<<ans;
}
### Prompt

In cpp, your task is to solve 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
/*input
18
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
*/
#include <bits/stdc++.h>
using namespace std;
int64_t n,ans;
string s;
map<pair<string,string>,int> l[19];
int main(){
	cin>>n>>s;
	for(int mask=0;mask<(1<<n);++mask){
		string red="",blue="";
		int num = __builtin_popcount(mask);
		for(int i=0;i<n;++i) 
			if((mask>>i)&1)
				red += s[i];
			else
				blue += s[i];
		reverse(blue.begin(),blue.end());
		++l[num][{red,blue}];
	}
	for(int mask=0;mask<(1<<n);++mask){
		string red="",blue="";
		int num = __builtin_popcount(mask);
		for(int i=0;i<n;++i) 
			if((mask>>i)&1)
				blue += s[2*n-1-i];
			else
				red += s[2*n-1-i];
		reverse(red.begin(),red.end());
		ans += l[num][{blue,red}];
	}
	cout<<ans;
}
```