Score : 400 points
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find ∑NK=1K×f(K).
Input is given from Standard Input in the following format:
N
Print the value ∑NK=1K×f(K).
4
23
We have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1×1+2×2+3×2+4×3=23.
100
26879
10000000
838627288460105
Watch out for overflows.