Notice that the solution equals the size of a team multiplied by the number of participating clubs, while a club participates if it has a member count divisible by the team size.

The easiest method to count the participating clubs for some team size ~S~ is iterating over all club sizes and counting clubs that have a size divisible by ~S~. The complexity of this algorithm is ~\mathcal O(N \times maxS)~, where ~maxS~ is the maximum club size, because the team size can be any value between ~1~ and ~maxS~, and counting for each team size takes ~\mathcal O(N)~.

We need a faster method of counting the competing clubs. Since club sizes are limited to ~2~ million, we can use an array ~a~ of size ~maxS~ to store, for each possible size, the number of clubs with that size.

In order to count the participating clubs for some team size ~d~, we need to compute the sum: ~a[d] + a[2d] + a[3d] + \dots~, which requires ~\frac{maxS}{d}~ steps.

Trying out all possibilities for ~d~ from ~1~ to ~maxS~ requires ~\frac{maxS}{1} + \frac{maxS}{2} + \dots + \frac{maxS}{maxS-1} + 1~ steps, which approximately equals ~maxS \ln maxS~. Thus, the complexity of this algorithm is ~\mathcal O(maxS \log maxS)~, which is sufficient to score all points.