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:

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?