2132：An easy problem

We once did a lot of recursional problem . I think some of them is easy for you and some if hard for you.
Now there is a very easy problem . I think you can AC it.
  We can define sum(n) as follow:
  if i can be divided exactly by 3 sum(i) = sum(i-1) + i*i*i;else sum(i) = sum(i-1) + i;
  Is it very easy ? Please begin to program to AC it..-_-

  The input file contains multilple cases.
  Every cases contain only ont line, every line contains a integer n (n<=100000).
  when n is a negative indicate the end of file.

  output the result sum(n).