Problem:
Score : 400 points

Problem Statement

For a positive integer X, let f(X) be the number of positive divisors of X.

Given a positive integer N, find \sum_{K=1}^N K\times f(K).

Constraints

1 \leq N \leq 10^7

Input

Input is given from Standard Input in the following format:

N

Output

Print the value \sum_{K=1}^N K\times f(K).

Sample Input 1

4

Sample Output 1

23

We have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1\times 1 + 2\times 2 + 3\times 2 + 4\times 3 =23.

Sample Input 2

100

Sample Output 2

26879

Sample Input 3

10000000

Sample Output 3

838627288460105

Watch out for overflows.

Solution in Ruby (2.7.1) (judge verdict: Runtime Error)

N = gets.to_i

FF = {}

SUM = {}

def divisor(num)
  ff = FF[num]
  return ff if ff
  result = []
  i = 1
  while i <= num do
    remainder = num % i
    if remainder == 0
      result << i
    end 
    i += 1
  end
  result_size =  result.size
  FF[num] = result_size
  result_size
end

def sum_f(x)
  sum_x = SUM[x]
  return sum_x if sum_x
  if x == 1
    result = divisor(x)
    SUM[x] = result
    return result
  end
  result = sum_f(x - 1) + x * divisor(x)
  SUM[x] = result
  result
end

ans = sum_f(N)

puts ans