#include <iostream>
#include <vector>
#include <string>
#include <map>
using namespace std;

#define int long long

signed main() {
	int N;
	cin >> N;
	string S;
	cin >> S;
	int ans = 0;
	map<string, int> mp;
	for (int msk = 0; msk < (1 << N); ++msk) {
		string red, blue;
		for (int i = 0; i < N; ++i) {
			if (msk & (1 << i))
				red += S[i];
			else
				blue += S[i];
		}
		++mp[red+' '+string(blue.rbegin(), blue.rend())];
	}
	for (int msk = 0; msk < (1 << N); ++msk) {
		string red, blue;
		for (int i = 0; i < N; ++i) {
			if (msk & (1 << i))
				red += S[N+i];
			else
				blue += S[N+i];
		}
		ans += mp[string(blue.rbegin(), blue.rend())+' '+red];
	}
	cout << ans << endl;
}

### 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 <iostream>
#include <vector>
#include <string>
#include <map>
using namespace std;

#define int long long

signed main() {
	int N;
	cin >> N;
	string S;
	cin >> S;
	int ans = 0;
	map<string, int> mp;
	for (int msk = 0; msk < (1 << N); ++msk) {
		string red, blue;
		for (int i = 0; i < N; ++i) {
			if (msk & (1 << i))
				red += S[i];
			else
				blue += S[i];
		}
		++mp[red+' '+string(blue.rbegin(), blue.rend())];
	}
	for (int msk = 0; msk < (1 << N); ++msk) {
		string red, blue;
		for (int i = 0; i < N; ++i) {
			if (msk & (1 << i))
				red += S[N+i];
			else
				blue += S[N+i];
		}
		ans += mp[string(blue.rbegin(), blue.rend())+' '+red];
	}
	cout << ans << endl;
}

```