Do two numbers contain unique factorials? Break two numbers up into their factorials; if they share any, return a falsey value. Otherwise, return a truthy value. (inspired by this recent question)

In other words, write each input number as the sum of factorials (of positive integers) in the greediest possible way; return a truthy value if no factorial appears in both representations, a falsey value otherwise.

Example

Given 20 and 49:

20 = 3! + 3! + 3! + 2!
49 = 4! + 4! + 1!


No factorial appears in both representations, so return a truthy value.

Given 32 and 132:

132 = 5! + 3! + 3!
 32 = 4! + 3! + 2!


3! appears in both representations, so return a falsey value.

I/O

Input and output can be through any standard means.

Input will always be two nonnegative integers; no upper bound on these integers other than what your language requires.

Output should be a truthy or falsey value. These values don't necessarily have to be consistent for different inputs, as long as every output is correctly truthy/falsey.

Test Cases

If one input is 0, the answer will always be truthy. Other truthy test cases:

{6, 3}, {4, 61}, {73, 2}, {12, 1}, {240, 2}, {5, 264}, {2, 91}, {673, 18},
 {3, 12}, {72, 10}, {121, 26}, {127, 746}


If both inputs are odd integers, or if both inputs are the same positive integer, then the output will always be falsey. Other falsey test cases:

{8, 5}, {7, 5}, {27, 47}, {53, 11}, {13, 123}, {75, 77}, {163, 160}, {148, 53},
 {225, 178}, {285, 169}, {39, 51}, {207, 334}, {153, 21}, {390, 128}, {506, 584},
 {626, 370}, {819, 354}


This is code-golf, so fewest bytes wins!
Jelly, 7 bytes

Æ!ṠḄ&amp;/¬


Try it online!

How it works

Æ!ṠḄ&amp;/¬  Main link. Argument: (x, y) (pair of integers)

Æ!       Convert x and y to factorial base.
  Ṡ      Apply the sign function to each digit.
   Ḅ     Unbinary; convert each resulting Boolean array from base 2 to integer.
    &amp;/   Reduce the resulting pair of integers by bitwise AND.
      ¬  Take the logical NOT of the result.