Planets Queries II - Walkthrough

Graph: $f(1) = 2, f(2) = 3, f(3) = 1, f(4) = 2$. Cycle: $1 \to 2 \to 3 \to 1$. Tail: $4$. Cycle length is $3$, tail node $4$ is distance $1$ from cycle. Query $(1, 3)$: both in cycle. Indices: $1$ at $0$, $3$ at $2$. Distance: $(2 - 0) \mod 3 = 2$.

From node $1$, go $2$ steps to reach node $3$. Query $(4, 3)$: $4$ in tail (distance $1$ to cycle at node $2$, index $1$). Jump to cycle, then $(2 - 1) \mod 3 = 1$. Total: $1 + 1 = 2$.