#include <iostream>
#include <string>
#include <unordered_map>
#include <utility>
#include <algorithm>
#define llint long long

using namespace std;

llint n;
string s;
unordered_map<string, llint> mp1, mp2;

void get(string s, unordered_map<string, llint> &mp)
{
	int N = 1 << n;
	string t, u;
	for(int i = 0; i < N; i++){
		t = u = "";
		for(int j = 0; j < n; j++){
			if(i & (1<<j)) t += s[j];
			else u += s[j];
		}
		mp[t+"#"+u]++;
	}
}

int main(void)
{
	cin >> n;
	cin >> s;
	
	get(s.substr(0, n), mp1);
	reverse(s.begin(), s.end());
	get(s.substr(0, n), mp2);
	
	llint ans = 0;
	for(auto it = mp1.begin(); it != mp1.end(); it++){
		ans += it->second * mp2[it->first];
	}
	cout << ans << endl;
	
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <iostream>
#include <string>
#include <unordered_map>
#include <utility>
#include <algorithm>
#define llint long long

using namespace std;

llint n;
string s;
unordered_map<string, llint> mp1, mp2;

void get(string s, unordered_map<string, llint> &mp)
{
	int N = 1 << n;
	string t, u;
	for(int i = 0; i < N; i++){
		t = u = "";
		for(int j = 0; j < n; j++){
			if(i & (1<<j)) t += s[j];
			else u += s[j];
		}
		mp[t+"#"+u]++;
	}
}

int main(void)
{
	cin >> n;
	cin >> s;
	
	get(s.substr(0, n), mp1);
	reverse(s.begin(), s.end());
	get(s.substr(0, n), mp2);
	
	llint ans = 0;
	for(auto it = mp1.begin(); it != mp1.end(); it++){
		ans += it->second * mp2[it->first];
	}
	cout << ans << endl;
	
	return 0;
}
```