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

typedef pair <string, string> P;

map <P, int> rec[2];

int main(){
	int n;
	string s;
	long long ans;
	cin >> n >> s;
	auto solve = [&](string str, int id){
		int len = str.size();
		string s1, s2;
		for (int i = 0; i < (1 << len); i++){
			s1.clear(); s2.clear();
			for (int j = 0; j < len; j++){
				if (i & (1 << j)) s1.push_back(str[j]);
				else s2.push_back(str[j]);
			}
			rec[id][{s1, s2}]++;
		}
	};
	solve(s.substr(0, n), 0);
	reverse(s.begin(), s.end());
	solve(s.substr(0, n), 1);
	ans = 0;
	for (auto it : rec[0])	
		ans += 1LL * it.second * rec[1][it.first];
	cout << ans << '\n';
	return 0;
}
### 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
#include <bits/stdc++.h>
using namespace std;

typedef pair <string, string> P;

map <P, int> rec[2];

int main(){
	int n;
	string s;
	long long ans;
	cin >> n >> s;
	auto solve = [&](string str, int id){
		int len = str.size();
		string s1, s2;
		for (int i = 0; i < (1 << len); i++){
			s1.clear(); s2.clear();
			for (int j = 0; j < len; j++){
				if (i & (1 << j)) s1.push_back(str[j]);
				else s2.push_back(str[j]);
			}
			rec[id][{s1, s2}]++;
		}
	};
	solve(s.substr(0, n), 0);
	reverse(s.begin(), s.end());
	solve(s.substr(0, n), 1);
	ans = 0;
	for (auto it : rec[0])	
		ans += 1LL * it.second * rec[1][it.first];
	cout << ans << '\n';
	return 0;
}
```