#include <bits/stdc++.h>

using namespace std;

int n;
long long res;
string s;
map< pair<string, string>, int> cnt;

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> s;
	for (int i = 0; i < (1 << n); ++i) {
		string A, B;
		for (int j = 0; j < n; ++j) {
			if (i >> j & 1) A.push_back(s[j]);
			else B.push_back(s[j]);
		}
		cnt[{A, B}]++;
	}
	for (int i = 0; i < (1 << n); ++i) {
		string A, B;
		for (int j = 0; j < n; ++j) {
			if (i >> j & 1) A.push_back(s[j + n]);
			else B.push_back(s[j + n]);
		}
		reverse(A.begin(), A.end());
		reverse(B.begin(), B.end());
		res += cnt[{A, B}];
	}
	cout << res;
}
### 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;

int n;
long long res;
string s;
map< pair<string, string>, int> cnt;

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> s;
	for (int i = 0; i < (1 << n); ++i) {
		string A, B;
		for (int j = 0; j < n; ++j) {
			if (i >> j & 1) A.push_back(s[j]);
			else B.push_back(s[j]);
		}
		cnt[{A, B}]++;
	}
	for (int i = 0; i < (1 << n); ++i) {
		string A, B;
		for (int j = 0; j < n; ++j) {
			if (i >> j & 1) A.push_back(s[j + n]);
			else B.push_back(s[j + n]);
		}
		reverse(A.begin(), A.end());
		reverse(B.begin(), B.end());
		res += cnt[{A, B}];
	}
	cout << res;
}
```