The algorithm iterates all outputs and collects the intersection of all
supported sizes. To fill the set initially, it assigns the current sizes if
the preliminary result is empty, but this condition can also be true if the
previous intersection resulted in an empty set. This is the case for instance
when using 3 monitors, of which the second one doesn't have any modes in common
with the first (e.g. rotation) and without this fix, it would return all sizes
of the third monitor instead of an empty set.

Avoid this by giving up early when there's no intersection left.

BUG: 442822