SQL query to get the owner's name, pet's name, and their age difference

Construct a query that outputs the owner's name under the alias keeper_name, the pet's name under the alias animal_name, and the difference between the owner's age and the pet's age as age_gap. Include all rows from the owners table; if the owner's name is missing (NULL), replace it with "N/A", and set age_gap to 0 in such cases. The sorting should be first by the owner's name in alphabetical order, then by the descending order of the age difference.

owners table:

ID	Name	Age	Pet_ID
1	Aubrey Little	20	1
2	Chett Crawfish	45	3
3	Jules Spinner	10	4
4	Magnus Burnsides	9	2
5	Veronica Dunn	8	NULL

pets table:

ID	Name	Age	Animal
1	Dr. Harris Bonkers	1	Rabbit
2	Moon	9	Dog
3	Ripley	7	Cat
4	Tom	2	Cat
5	Maisie	10	Dog