Prove that (p → q) → ((r ∨ p) → (r ∨ q)) is a tautology without using truth table

I can't find a proper formula for this considering it's almost exclusively made up of implications. Can somebody help me?

just I want to know how to simplify it.

I tried this but still no correct answerenter image description here

If r is false, then ((r ∨ p) → (r ∨ q)) is just (p → q).

If r is true, then ((r ∨ p) → (r ∨ q)) is just (true → true).