output_description: Print the value \sum_{K=1}^N K\times f(K).
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: import sys
  import numpy as np
  
  read = sys.stdin.buffer.read
  readline = sys.stdin.buffer.readline
  readlines = sys.stdin.buffer.readlines
  
  
  def prime_table(N):
      is_prime = np.zeros(N, np.int64)
      is_prime[2:3] = 1
      is_prime[3::2] = 1
      for p in range(3, N, 2):
          if p * p >= N:
              break
          if is_prime[p]:
              is_prime[p * p :: p + p] = 0
      return is_prime, np.where(is_prime)[0]
  
  
  def main(N, primes):
      div = np.ones(N + 1, dtype=np.int64)
      for p in primes:
          for i in range(N // p + 1):
              div[p * i] += div[i]
      div *= np.arange(N + 1)
      return div.sum()
  
  
  if sys.argv[-1] == "ONLINE_JUDGE":
      import numba
      from numba.pycc import CC
  
      i8 = numba.int64
      cc = CC("my_module")
  
      def cc_export(f, signature):
          cc.export(f.__name__, signature)(f)
          return numba.njit(f)
  
      main = cc_export(main, (i8, i8[:]))
      cc.compile()
  
  from my_module import main
  
  N = int(read())
  is_prime, primes = prime_table(N + 1)
  print(main(N, primes))
  
problem_description: 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).
samples: [{"input": "4", "output": "23\n    \n\nWe have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1\\times 1 +\n2\\times 2 + 3\\times 2 + 4\\times 3 =23.\n\n* * *"}, {"input": "100", "output": "26879\n    \n\n* * *"}, {"input": "10000000", "output": "838627288460105\n    \n\nWatch out for overflows."}]