#include <bits/stdc++.h>
using namespace std;

typedef pair<string, string> pss;
map<pss, int> mp;


long long ans, n;
string s;
int main() {
	cin >>n >> s;
	int mid = (int)s.size() / 2, mid2 = (int)s.size() - mid;
	for (int i = 0; i < (1<<mid); i++) {
		string s1, t1;
		for (int j = 0; j < mid; j++) {
			if ((i>>j) & 1) 
				s1 += s[j];
			else 
				t1 += s[j];
		}
		reverse(t1.begin(), t1.end());
		mp[pss(s1, t1)]++;
	}
	for (int i = 0; i < (1<<mid2); i++) {
		string s1, t1;
		for (int j = 0; j < mid2; j++) {
			if ((i>>j) & 1) 
				s1 += s[j+mid];
			else 
				t1 += s[j+mid];
		}
		reverse(s1.begin(), s1.end());
		ans += mp[pss(s1, t1)];
		
	}
	cout << ans;
}
### Prompt

Your challenge is to write 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;

typedef pair<string, string> pss;
map<pss, int> mp;


long long ans, n;
string s;
int main() {
	cin >>n >> s;
	int mid = (int)s.size() / 2, mid2 = (int)s.size() - mid;
	for (int i = 0; i < (1<<mid); i++) {
		string s1, t1;
		for (int j = 0; j < mid; j++) {
			if ((i>>j) & 1) 
				s1 += s[j];
			else 
				t1 += s[j];
		}
		reverse(t1.begin(), t1.end());
		mp[pss(s1, t1)]++;
	}
	for (int i = 0; i < (1<<mid2); i++) {
		string s1, t1;
		for (int j = 0; j < mid2; j++) {
			if ((i>>j) & 1) 
				s1 += s[j+mid];
			else 
				t1 += s[j+mid];
		}
		reverse(s1.begin(), s1.end());
		ans += mp[pss(s1, t1)];
		
	}
	cout << ans;
}
```