Hints: Minimum Path Cover
A vertical path goes parent $\to$ child $\to$ $\cdots$ along the tree. A path is good if the GCD of values on it is $> 1$. You must partition every node into disjoint good vertical paths.
If a node’s value is prime and shares no factor with its parent’s value, what must happen to that node in any feasible cover?
Answer to Hint 1: Such a node cannot extend a path from its parent with GCD $> 1$ unless the parent already carries a compatible factor. Often it must start its own singleton path (still “good” because a single value $> 1$ has GCD equal to itself).
The challenge is merging children’s paths upward when a common divisor appears.
Answer to Hint 2: The tree is revealed online from node $n$ down to $1$. When you attach children to the current node $v$ with value $a_v$, you only know future structure through what you have already computed for subtrees of larger labels.
Think of each subtree as returning not only its answer $f(S(u))$ but a compact description of which GCD residues can still continue upward through $v$.
Answer to Hint 3: For each node, maintain a multiset (or list) of integers representing “active” GCDs achievable at the bottom of an ongoing vertical path that ends ready to combine at this node. When processing a child’s contribution $g$, combine with $a_v$ via $\gcd(g, a_v)$.
If the GCD becomes $1$, that continuation dies; otherwise split factors using $\gcd$ structure (divide out common parts) so the state stays manageable.
Answer to Hint 4: Merge all children: sum their path counts, then repeatedly feed child states into a combining routine that updates the parent’s multiset. If after processing all children the multiset is empty, you must start a fresh path at $v$ (increment answer by $1$) and insert $a_v$ as a new state.
Otherwise, surviving states mean some vertical paths can continue higher without opening a new path at $v$.
Answer to Hint 5: The gcd splitting in the reference solution keeps numbers by repeatedly factoring out $\gcd(x,y)$ from a running value and pushing reduced pieces—this avoids storing exponentially many paths explicitly.
Flush output after each printed line; the interactor feeds the next node only after reading your answer for $f(S(\text{next}))$.