#include <bits/stdc++.h>

using namespace std;

int main() {
	int n;
	string s;
	cin >> n >> s;
	map<pair<string, string>, long> m;
	for (int i = 0; i < 1 << n; i++) {
		string t1, t2;
		for (int j = 0; j < n; j++) {
			if (i & 1 << j) t1 += s[j];
			else t2 += s[j];
		}
		m[make_pair(t1, t2)]++;
	}
	long r = 0;
	for (int i = 0; i < 1 << n; i++) {
		string t1, t2;
		for (int j = 0; j < n; j++) {
			if (i & 1 << j) t1 += s[n * 2 - j - 1];
			else t2 += s[n * 2 - j - 1];
		}
		auto it = m.find(make_pair(t2, t1));
		if (it != m.end())
			r += it->second;
	}
	cout << r << endl;
	return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 main() {
	int n;
	string s;
	cin >> n >> s;
	map<pair<string, string>, long> m;
	for (int i = 0; i < 1 << n; i++) {
		string t1, t2;
		for (int j = 0; j < n; j++) {
			if (i & 1 << j) t1 += s[j];
			else t2 += s[j];
		}
		m[make_pair(t1, t2)]++;
	}
	long r = 0;
	for (int i = 0; i < 1 << n; i++) {
		string t1, t2;
		for (int j = 0; j < n; j++) {
			if (i & 1 << j) t1 += s[n * 2 - j - 1];
			else t2 += s[n * 2 - j - 1];
		}
		auto it = m.find(make_pair(t2, t1));
		if (it != m.end())
			r += it->second;
	}
	cout << r << endl;
	return 0;
}

```