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1 [made Community Wiki]

To change up the nature of the responses some, IMO a good theorem to think about is the Kan-Thurston theorem. It states that given any space $X$ you can find a $K(\pi, 1)$ space $Y$ and a map $f : Y \to X$ inducing isomorphisms $f_* : H_i Y \to H_i X$, $f^* : H^i X \to H^i Y$ for all coefficients (it can be souped-up to allow local coefficients) and all $i$. The map $\pi_1 Y \to \pi_1 X$ is onto.

So from the point of view of cohomology algebras with Steenrod operations, these spaces are the same. One way to "spin" this would be to say the fundamental group is a far stronger invariant than anything (co)homological.