#include <bits/stdc++.h>

using namespace std;

int n;
string s;

long long ans;
map<pair<string,string>,long long> f,r;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n >> s;

	string rs,fs;

	fs = s.substr(0,n);
	rs = s.substr(n,n);

	for (int i = 0;i < (1<<n);i++){
		string s1="",s2="",s3="",s4="";
		for (int j = 0;j < n;j++){
			if(i&(1<<j))s1+=fs[j];
			else s2+=fs[j];
			if(i&(1<<j))s3+=rs[j];
			else s4+=rs[j];	
		}
		reverse(s2.begin(),s2.end());
		reverse(s4.begin(),s4.end());
		f[make_pair(s1,s2)]++;
		r[make_pair(s4,s3)]++;
	}

	for (auto x : f){
		ans += x.second*r[x.first];
	}

	cout << ans << endl;
	
	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>

using namespace std;

int n;
string s;

long long ans;
map<pair<string,string>,long long> f,r;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n >> s;

	string rs,fs;

	fs = s.substr(0,n);
	rs = s.substr(n,n);

	for (int i = 0;i < (1<<n);i++){
		string s1="",s2="",s3="",s4="";
		for (int j = 0;j < n;j++){
			if(i&(1<<j))s1+=fs[j];
			else s2+=fs[j];
			if(i&(1<<j))s3+=rs[j];
			else s4+=rs[j];	
		}
		reverse(s2.begin(),s2.end());
		reverse(s4.begin(),s4.end());
		f[make_pair(s1,s2)]++;
		r[make_pair(s4,s3)]++;
	}

	for (auto x : f){
		ans += x.second*r[x.first];
	}

	cout << ans << endl;
	
	return 0;

}
```