First, make a list containing all the chocolates and all the boxes, and sort it in the decreasing order of their widths. If there is a chocolate and a box with the same width, order the box first.

Prepare an empty integer sequence \(S=()\). Inspect each element as follows.

• If the inspected element is a box \((C_i,D_i)\)
  Add \(D_i\) to \(S\).

• If the inspected element is a chocolate \((A_i,B_i)\)
  Remove the smallest element of \(S\) greater than or equal to \(B_i\). If there is no element to remove, the answer is No.

If all the elements were successfully inspected, then the answer is Yes.

The naive implementation of the solution above is as slow as \(O(NM)\), but we can use std::multiset for \(S\) or use a segment tree after compressing coordinates so that it can be solved in a total of \(O((N+M)\log (N+M))\) time, which is fast enough.