#include <bits/stdc++.h>
using namespace std;

int n;
long long ans;
string s;
map < pair < string, string >, long long > m;

signed main () {
	cin >> n >> s;
	for (int i = 0; i < (1 << n); ++i) {
		string red, blue;
		for (int j = 0; j < n; ++j) {
			if (i & (1 << j)) 
				red += s[j];
			else 
				blue += s[j];
		}
		reverse(blue.begin(), blue.end());
		++m[make_pair(red, blue)];
	}
	for (int i = 0; i < (1 << n); ++i) {
		string red, blue;
		for (int j = 0; j < n; ++j) {
			if (i & (1 << j)) 
				red += s[n + j];
			else 
				blue += s[n + j];
		}
		reverse(blue.begin(), blue.end());
		ans += m[make_pair(blue, red)];
	}
	cout << ans << '\n';
}

### Prompt

Please formulate a cpp solution 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;

int n;
long long ans;
string s;
map < pair < string, string >, long long > m;

signed main () {
	cin >> n >> s;
	for (int i = 0; i < (1 << n); ++i) {
		string red, blue;
		for (int j = 0; j < n; ++j) {
			if (i & (1 << j)) 
				red += s[j];
			else 
				blue += s[j];
		}
		reverse(blue.begin(), blue.end());
		++m[make_pair(red, blue)];
	}
	for (int i = 0; i < (1 << n); ++i) {
		string red, blue;
		for (int j = 0; j < n; ++j) {
			if (i & (1 << j)) 
				red += s[n + j];
			else 
				blue += s[n + j];
		}
		reverse(blue.begin(), blue.end());
		ans += m[make_pair(blue, red)];
	}
	cout << ans << '\n';
}

```