Let p and q be prime numbers and 𝜙 Euler’s totient function. Is there an efficient way of computing 𝜙(𝜙(p*q)) = 𝜙(𝜙((p-1)(q-1)), that is not simply based on factoring (p-1) and (q-1)?

Obviously, if p and q do not equal two, (p-1) and (q-1) are even and consequently their prime factorization is entirely different from the prime factorization of p and q. Therefore I assume that no such shortcut exists.

Do I overlook something?