Hints: Oriented Journey

There are $2^{n-1}$ possible values of $s$, and Mashiro only sees a directed tree on $n$ labeled vertices. How many bits of information must the orientation encode in the worst case?

Answer to Hint 1: You need $n-1$ independent binary choices — exactly the number of edges. Try fixing vertex $1$ as a root and using each edge’s direction to reveal one bit of $s-1$ in a fixed labeling-independent way.

Answer to Hint 2: Assign each vertex $i \in \{2,\ldots,n\}$ a target bit $X_i$, equal to bit $(i-2)$ of $s-1$. When an edge connects $u$ and $v$, orient it so that walking from vertex $1$ along the tree xors the bits along directed edges to recover $X$ at each vertex (use a consistent rule based on whether $X_u = X_v$).

Answer to Hint 3: Model solution: for each undirected edge with $u < v$, if $X_u = X_v$ output $(u,v)$, else output $(v,u)$. Root the tree at $1$, treat each directed edge as giving xor $0$ or $1$ from parent to child, BFS to recover all $X_i$, then reconstruct $s = 1 + \sum_{i=2}^{n} X_i \cdot 2^{i-2}$.

Answer to Hint 4: The second run may shuffle test cases; you only need to decode $s$ from each directed tree independently — no cross-test state.

Answer to Hint 5: Remember to flush after each interactive print (cout << ... << endl or cout.flush()), and exit immediately on reading -1.