#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n;
char s[100];
ll ans;
map<pair<unsigned int, unsigned int>, int> M;
int main() {
	cin >> n;
	cin >> s;
	for (int i = 0; i < (1 << n); ++i) {
		unsigned int t1 = 0, t2 = 0;
		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				t1 *= 13131;
				t1 += s[j];
			} else {
				t2 *= 13131;
				t2 += s[j];
			}
		}
		M[make_pair(t1, t2)]++;
	}
	reverse(s, s + 2 * n);
	for (int i = 0; i < (1 << n); ++i) {
		unsigned int t1 = 0, t2 = 0;
		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				t1 *= 13131;
				t1 += s[j];
			} else {
				t2 *= 13131;
				t2 += s[j];
			}
		}
		ans += M[make_pair(t2, t1)];
	}
	cout << ans;
}
### 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<bits/stdc++.h>
#define ll long long
using namespace std;
int n;
char s[100];
ll ans;
map<pair<unsigned int, unsigned int>, int> M;
int main() {
	cin >> n;
	cin >> s;
	for (int i = 0; i < (1 << n); ++i) {
		unsigned int t1 = 0, t2 = 0;
		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				t1 *= 13131;
				t1 += s[j];
			} else {
				t2 *= 13131;
				t2 += s[j];
			}
		}
		M[make_pair(t1, t2)]++;
	}
	reverse(s, s + 2 * n);
	for (int i = 0; i < (1 << n); ++i) {
		unsigned int t1 = 0, t2 = 0;
		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				t1 *= 13131;
				t1 += s[j];
			} else {
				t2 *= 13131;
				t2 += s[j];
			}
		}
		ans += M[make_pair(t2, t1)];
	}
	cout << ans;
}
```