Each of the six faces on a cube has a different digit (0 to 9) written on it;
the same is done to a second cube.
By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.

For example, the square number 64 could be formed:

~~~
  +-------+     +-------+
 /       /|    /       /|
+-------+ |   +-------+ |
|       | |   |       | |
|   6   | +   |   4   | +
|       |/    |       |/
+-------+     +-------+
~~~

In fact, by carefully choosing the digits on both cubes it is possible to display
all of the square numbers below one-hundred: `01`, `04`, `09`, `16`, `25`, `36`, `49`, `64`, and `81`.

For example, one way this can be achieved is by placing `{0, 5, 6, 7, 8, 9}` on one cube 
and `{1, 2, 3, 4, 8, 9}` on the other cube.

However, for this problem we shall allow the `6` or `9` to be turned upside-down 
so that an arrangement like `{0, 5, 6, 7, 8, 9}` and `{1, 2, 3, 4, 6, 7}` allows 
for all nine square numbers to be displayed; otherwise it would be impossible to obtain `09`.

In determining a distinct arrangement we are interested in the digits on each cube, not the order.

 - `{1, 2, 3, 4, 5, 6}` is equivalent to `{3, 6, 4, 1, 2, 5}`
 - `{1, 2, 3, 4, 5, 6}` is distinct from `{1, 2, 3, 4, 5, 9}`

But because we are allowing `6` and `9` to be reversed, the two distinct sets in 
the last example both represent the extended set `{1, 2, 3, 4, 5, 6, 9}` for the 
purpose of forming 2-digit numbers.

How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?