Let's assume a fixed ~K~ (the number of consecutive numbers) and denote the number of happy numbers in the set ~\{i, i+1, \dots, i+K-1\}~ with ~f(i)~. It is clear that ~|f(i)-f(i+1)| \le 1~.

Additionally, we can use brute force to find the first ~150~ consecutive composite numbers. If they begin with the number ~j~, then it will hold ~f(j) = 0~ for each ~K \le 150~.

Lemma: Let ~a~ and ~b~ be integers and let ~f(a) \le L \le f(b)~. Then there exists ~x~ from the interval ~[a, b]~ such that ~f(x) = L~.

The proof of this lemma is simple and is left as an exercise to the reader.

Now, let's work with the numbers ~K, L, M~ from the task. By applying the lemma to ~a = j~ and ~b = 1~, we know that a solution must exist in the interval ~[1, j]~. However, iterating over the entire interval would be too slow.

We use the binary search technique. Let's assume that we know for sure that our solution is in the interval ~[a, b]~. Let ~c = (a+b)/2~. Then we can apply the lemma to one of the intervals ~[a, c]~ or ~[c, b]~ and solve the task recursively.

The total time complexity of this algorithm is ~\mathcal O(Q \log j)~.