#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int n; string s;
	cin >> n >> s;
	vector<pair<string, string> > fw, bk;
	for (int i = 0; i < 1 << n; ++i) {
		string fwa, fwb, bka, bkb;
		for (int j = 0; j < n; ++j) {
			if ((i >> j) & 1) {
				fwa += s[j];
				bka += s[2 * n - j - 1];
			}
			else {
				fwb += s[j];
				bkb += s[2 * n - j - 1];
			}
		}
		fw.push_back(make_pair(fwa, fwb));
		bk.push_back(make_pair(bkb, bka));
	}
	sort(fw.begin(), fw.end());
	sort(bk.begin(), bk.end());
	long long ans = 0;
	for (pair<string, string> i : fw) {
		ans += upper_bound(bk.begin(), bk.end(), i) - lower_bound(bk.begin(), bk.end(), i);
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Create a solution in CPP for 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 <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int n; string s;
	cin >> n >> s;
	vector<pair<string, string> > fw, bk;
	for (int i = 0; i < 1 << n; ++i) {
		string fwa, fwb, bka, bkb;
		for (int j = 0; j < n; ++j) {
			if ((i >> j) & 1) {
				fwa += s[j];
				bka += s[2 * n - j - 1];
			}
			else {
				fwb += s[j];
				bkb += s[2 * n - j - 1];
			}
		}
		fw.push_back(make_pair(fwa, fwb));
		bk.push_back(make_pair(bkb, bka));
	}
	sort(fw.begin(), fw.end());
	sort(bk.begin(), bk.end());
	long long ans = 0;
	for (pair<string, string> i : fw) {
		ans += upper_bound(bk.begin(), bk.end(), i) - lower_bound(bk.begin(), bk.end(), i);
	}
	cout << ans << endl;
	return 0;
}
```