A closed subset of Baire space not Medvedev equivalent to any closed set of Cantor space

For mass problems $P,Q\subseteq {\mathbb{N}}^{\mathbb{N}}$ (Baire space), $P$ is Medvedev reducible to $Q$ ($P{\le }_{s}Q$) if for some Turing funcional $\mathrm{\Phi }$, $\mathrm{\Phi }(Q)\subseteq P$, and Medvedev equivalent to $Q$ if also $Q{\le }_{s}P$. Shafer asked if every closed problem $P$ is Medvedev equivalent to a closed problem $Q$ with $Q\subseteq 2^{\mathbb{N}}$ (Cantor space). We show that this is not the case.