As a general observation, we can note that keys are essentially interchangeable. If we're "using" \(x\) mailboxes (that is, there are \(x\) non-empty mailboxes that we intend for the thief to gain access to), the only requirement is that a total of at least \(x-1\) keys are present somewhere amongst them. It's then always possible to assign the keys' identities and select an initial break-in mailbox such that the thief will eventually access all \(x\) mailboxes.

Now, we'll begin by sorting the mailboxes in order of capacity, and the packages in order of size. We'll then binary search on \([0, \min(N, M)]\) for the maximum number \(v\) of packages which can be obtained, which will be our answer.

For a given value of \(v\), we'll begin by reducing the set of \(N+M\) mailboxes/packages to only the relevant ones which we'll intend to use. We should attempt to use the smallest \(v\) packages, ignoring the rest. We'll need to use at least \(v\) mailboxes, which might as well be the largest \(v\) of them. If there are additional mailboxes capable of fitting at least one key (having capacities no less than \(X\)), then we should intend on using all of them as well (even if some may prove unnecessary), and we should **ignore any remaining mailboxes**. It can't be better to use fewer mailboxes, as using a mailbox capable of fitting a key can never be detrimental (as it can at least fit a key allowing use of another mailbox in its place). It also can't be better to use more mailboxes, as using an *additional* mailbox with capacity less than \(X\) (with the intent of placing a package with size less than \(X\) in it) cannot be more beneficial than instead replacing an existing key in some larger mailbox not yet containing any package with this small package — at least one such mailbox must exist in that scenario.

We'll then combine the relevant mailboxes/packages into a single list sorted in non-increasing order of capacity/size, and iterate over them in that order. Along the way, we'll maintain an ordered multiset of mailbox capacities which have been seen but not yet used, ordered by their capacities modulo \(X\) (initially empty), as well as the total number \(k\) of keys which can fit within the already-seen mailboxes (initially \(0\)).

- When a mailbox with capacity \(c\) is encountered, we'll insert \(c\) into our multiset and increase \(k\) by \(\lfloor c/X \rfloor\).
- When a package with size \(p\) is encountered, it must be placed into some mailbox in our multiset. If the multiset is empty, then the given value of \(v\) must be invalid. Otherwise, we should choose a mailbox capacity \(c\) for which \(c \text{ \% } X \ge p \text{ \% } X\) and \(c \text{ \% } X\) is minimized — or, if there's no such capacity, then instead just a capacity \(c\) for which \(c \text{ \% } X\) is minimized (where \(\text{\%}\) is the modulo operator). This greedy strategy is based on the fact that the package will reduce the number of keys which can fit in the mailbox by \(\lfloor p/X \rfloor\) if \(c \text{ \% } X \ge p \text{ \% } X\), and by \(\lfloor p/X \rfloor + 1\) otherwise. Either way, once the optimal \(c\) has been selected, we'll remove it from our multiset and decrease \(k\) accordingly.

By the end of this process, the given value of \(v\) must be valid if and only if \(k+1\) is greater than or equal to the number of mailboxes used.

The total time complexity of this solution is \(O((N + M) (log(N) + log(M))^2)\).