Hints: Deconstruction Tree

Do E0 first.
For the full problem, assume you already know which vertices are potentially reachable and now need to count distinct sets.

Answer to Hint 1: Keep the same root-at-$n$ preprocessing:

• $\mathrm{bel}[v]$: branch id (which child of $n$),
• $\mathrm{sub}[v]$: max label in rooted subtree of $v$ away from $n$.

Only champions with $\mathrm{sub}[v] < v$ can contribute to counting.

Answer to Hint 2: Let $s$ be the maximum-index initial leaf. Round 1 is fixed, so start with:

$$ \mathrm{dp}[s] = 1,\quad \mathrm{pre}[s] = 1. $$

Answer to Hint 3: Define DP meaning clearly:

• $\mathrm{dp}[i]$ = number of new distinct set-patterns introduced when processing label $i$.
• $\mathrm{pre}[i] = \sum_{j=s}^{i}\mathrm{dp}[j]$ = total patterns available up to label $i$.

Answer to Hint 4: Transition for $i > s$:

• if $\mathrm{sub}[i] \ge i$, then $\mathrm{dp}[i] = 0$;
• else $$ \mathrm{dp}[i] = \mathrm{pre}[i - 1] - \mathrm{pre}[\mathrm{sub}[i]]. $$
• always update $$ \mathrm{pre}[i] = \mathrm{pre}[i - 1] + \mathrm{dp}[i]. $$

Answer to Hint 5: Why subtract $\mathrm{pre}[\mathrm{sub}[i]]$?

• $\mathrm{pre}[i - 1]$ counts everything formed so far.
• Patterns already fixed by the same-branch frontier at $\mathrm{sub}[i]$ are not new.
• Removing those duplicates leaves exactly new patterns created at $i$.

Answer to Hint 6: Not all champions are summed in the final answer.
Let $B = \mathrm{bel}[n - 1]$ and let $\mathrm{other}$ be the largest $i < n$ with $\mathrm{bel}[i] \ne B$.

Only indices $i > \max(s, \mathrm{other})$ on branch $B$ remain free at the end.

Answer to Hint 7: Final formula:

$$ \sum_{\substack{s < i < n \\ \mathrm{bel}[i] = \mathrm{bel}[n - 1] \\ i > \max(s, \mathrm{other})}} \mathrm{dp}[i]. $$

Answer to Hint 8: In sample 3 and sample 4, this sum collapses to one term:

• sample 3: $\mathrm{dp}[6] = 3$;
• sample 4: $\mathrm{dp}[9] = 13$.

That is why those outputs are $3$ and $13$.

Answer to Hint 9: Full algorithm per test case:

1. If the max-index leaf is $n$, print $1$.
2. DFS from $n$: fill $\mathrm{bel}[\cdot]$ and $\mathrm{sub}[\cdot]$.
3. Sweep $i = s, s + 1, \ldots, n - 1$ to compute $\mathrm{dp}$ and $\mathrm{pre}$.
4. Find $\mathrm{other}$, then sum $\mathrm{dp}[i]$ on branch $\mathrm{bel}[n - 1]$ as above.

Time $O(n)$, memory $O(n)$. The mod $998\,244\,353$ wrapper in the model solution is only for the type system — the combinatorics are ordinary integer counts that fit in the recurrence.