Antimagic Square

Topic Description
Ancient Chinese books have long been recorded.

2 9 4
7 5 3
6 1 8

This is a third-order magic square. Each row, column and diagonal number adds up equally.

Let's consider the opposite question.
Can you fill in the nine palaces with numbers 1 to 9?
Make: Each row, column, and diagonal number is not equal to each other?

That should be possible.
For example:
9 1 2
8 4 3
7 5 6

Your task is to search all third-order anti-magic squares. And count out how many kinds there are altogether.
Rotation or mirroring is the same.

For example:
9 1 2
8 4 3
7 5 6

7 8 9
5 4 1
6 3 2

2 1 9
3 4 8
6 5 7

Wait counts as the same situation.

Please submit the total number of third-order anti-magic squares. This is an integer. Don't fill in any redundancy.

public class Main {
	static int ans = 0;
 
	public static void main(String[] args) {
		int[] a = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
		f(a, 0, a.length - 1);
		System.out.println(ans / 8);
	}
 
	public static void f(int[] a, int start, int end) {
		if (start == 6) {
			int sum1 = a[0] + a[1] + a[2];
			int sum2 = a[3] + a[4] + a[5];
			if (sum1 == sum2)
				return;
		} else if (start == 7) {
			int sum1 = a[0] + a[1] + a[2];
			int sum2 = a[3] + a[4] + a[5];
			int sum3 = a[2] + a[4] + a[6];
			int sum4 = a[0] + a[3] + a[6];
			if (sum1 == sum2 || sum1 == sum3 || sum1 == sum4 || sum2 == sum3 || sum2 == sum4 || sum3 == sum4)
				return;
		} else if (start == 8) {
			int sum1 = a[0] + a[1] + a[2];
			int sum2 = a[3] + a[4] + a[5];
			int sum3 = a[2] + a[4] + a[6];
			int sum4 = a[0] + a[3] + a[6];
			int sum5 = a[1] + a[4] + a[7];
			int sum6 = a[2] + a[5] + a[8];
			int sum7 = a[0] + a[4] + a[8];
			int sum8 = a[6] + a[7] + a[8];
			if (sum1 == sum2 || sum1 == sum3 || sum1 == sum4 || sum1 == sum5 || sum1 == sum6 || sum1 == sum7
					|| sum1 == sum8 || sum2 == sum3 || sum2 == sum4 || sum2 == sum5 || sum2 == sum6 || sum2 == sum7
					|| sum2 == sum8 || sum3 == sum4 || sum3 == sum5 || sum3 == sum6 || sum3 == sum7 || sum3 == sum8
					|| sum4 == sum5 || sum4 == sum6 || sum4 == sum7 || sum4 == sum8 || sum5 == sum6 || sum5 == sum7
					|| sum5 == sum8 || sum6 == sum7 || sum6 == sum8 || sum7 == sum8)
				return;
			ans++;
			return;
		}
		for (int i = start; i <= end; i++) {
			{
				int temp = a[start];
				a[start] = a[i];
				a[i] = temp;
			}
			f(a, start + 1, end);
			{
				int temp = a[start];
				a[start] = a[i];
				a[i] = temp;
			}
		}
	}
}
 
// Answer: 3120