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

int main() {
	long long N, ANS = 0;
	string S, X, Y;
	map<pair<string, string>, long long> mp;
	cin >> N >> S;
	X = S.substr(0, N);
	Y = S.substr(N, N);
	reverse(Y.begin(), Y.end());
	for (long long i = 0; i < 1LL << N; i++) {
		long long I = i;
		string R = "", B = "";
		for (int j = 0; j < N; j++) {
			if (I % 2 == 0) R += X[j];
			else B += X[j];
			I /= 2;
		}
		mp[{R, B}]++;
	}
	for (long long i = 0; i < 1LL << N; i++) {
		long long I = i;
		string R = "", B = "";
		for (int j = 0; j < N; j++) {
			if (I % 2 == 0) R += Y[j];
			else B += Y[j];
			I /= 2;
		}
		ANS += mp[{R, B}];
	}
	cout << ANS << endl;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 main() {
	long long N, ANS = 0;
	string S, X, Y;
	map<pair<string, string>, long long> mp;
	cin >> N >> S;
	X = S.substr(0, N);
	Y = S.substr(N, N);
	reverse(Y.begin(), Y.end());
	for (long long i = 0; i < 1LL << N; i++) {
		long long I = i;
		string R = "", B = "";
		for (int j = 0; j < N; j++) {
			if (I % 2 == 0) R += X[j];
			else B += X[j];
			I /= 2;
		}
		mp[{R, B}]++;
	}
	for (long long i = 0; i < 1LL << N; i++) {
		long long I = i;
		string R = "", B = "";
		for (int j = 0; j < N; j++) {
			if (I % 2 == 0) R += Y[j];
			else B += Y[j];
			I /= 2;
		}
		ANS += mp[{R, B}];
	}
	cout << ANS << endl;
}
```