UVA 11762

Suppose we want to calculate E(N) where N is a number,k is the number of it’s distinct prime factors and p is the number of primes < n Normal rule, E(N)=(1/p)*(E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p)*(1 + E(N))

I don't understand the editorial part of this problem.... Below part is the editorial part..

E(N)=1+(1/p)*(E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p) * E(N)

-Why here add 1 in the first? Why not add 1 in the failure part?