#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct ss {
	string a, b;
	bool operator < (ss o) const {
		return tie(a, b) < tie(o.a, o.b);
	}
};

int N;
string str;
map<ss, ll> cnt;

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> N;
	cin >> str;
	
	for (int mask = 0; mask < (1<<N); mask++) {
		string x, y;
		for (int i = 0; i < N; i++) {
			if (mask & (1<<i)) x += str[i];
			else y += str[i];
		}
		cnt[{x, y}]++;
	}
	
	ll ans = 0;
	for (int mask = 0; mask < (1<<N); mask++) {
		string x, y;
		for (int i = 0; i < N; i++) {
			if (mask & (1<<i)) x += str[2*N-1-i];
			else y += str[2*N-1-i];
		}
		ans += cnt[{x, y}];
	}
	cout << ans << "\n";
	
	return 0;
}

### Prompt

Please create a solution in Cpp 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 long long ll;
struct ss {
	string a, b;
	bool operator < (ss o) const {
		return tie(a, b) < tie(o.a, o.b);
	}
};

int N;
string str;
map<ss, ll> cnt;

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> N;
	cin >> str;
	
	for (int mask = 0; mask < (1<<N); mask++) {
		string x, y;
		for (int i = 0; i < N; i++) {
			if (mask & (1<<i)) x += str[i];
			else y += str[i];
		}
		cnt[{x, y}]++;
	}
	
	ll ans = 0;
	for (int mask = 0; mask < (1<<N); mask++) {
		string x, y;
		for (int i = 0; i < N; i++) {
			if (mask & (1<<i)) x += str[2*N-1-i];
			else y += str[2*N-1-i];
		}
		ans += cnt[{x, y}];
	}
	cout << ans << "\n";
	
	return 0;
}

```