Odd-order and normal rank two for all primes implies Sylow tower

Statement
Suppose $$G$$ is a finite group of odd order. Further, suppose that for every prime $$p$$ dividing the order of $$G$$, and every $$p$$-Sylow subgroup $$P$$ of $$G$$, the normal rank of $$P$$ is at most two. Then, $$G$$ has a Sylow tower.

Breakdown for even order
The result fails if we allow groups of even order. The smallest counterexample is the symmetric group of degree four, where both Sylow subgroups have normal rank at most two, but the group does not possess a Sylow tower.