Output #1
3 4
We have an interval $[1,6]$, and queries for the intervals $[2,3],[3,4],[2,4],[3,3]$. Here, if you use the binary tree shown in the figure below,
* for the first query, nodes $1,2,3,4,5$ are investigated, and the maximum depth among these is $2$ for nodes $4,5$;
* for the second query, nodes $1,2,3,4,5,6,7,8,9$ are investigated, and the maximum depth among these is $3$ for nodes $8,9$;
* for the third query, nodes $1,2,3,4,5,6,7$ are investigated, and the maximum depth among these is $2$ for nodes $4,5,6,7$;
* for the fourth query, nodes $1,2,3,4,5,8,9$ are investigated, and the maximum depth among these is $3$ for nodes $8,9$.
Therefore, in this case, $d=3$, and nodes of depth $3$ were investigated four times ー nodes $8,9$ in the second query and nodes $8,9$ in the fourth query ー so $c=4$. In fact, this is an optimal method.
(In the figure, the upper row shows the node number, and the lower row shows the interval of each node.)
