Prove or disprove: If $G$ is a CFG in Chomsky normal form, then for any string $w \in L(G)$ of length $n\geq 1$ then exactly $2n-1$ steps are required for any derivation of $w$.
I'm stuck at where to start with this proof... When i see the $2n-1$ it leads me to believe that I should utilize the pigeon hole principle. However, i'm also thinking it would be more thorough to prove this by doing a proof by construction. Any help is appreciated.