message: Codeforces 1520B solution
  Codeforces ordinary numbers problem solution
diff: @@ -22,7 +22,7 @@ units place : 10^0
  tens place : 10^1
  hundreds place : 10^2 and so on;
  Here we observe a pattern that any ordinary number in the form (nnnnnn..) having say m digits
  -can be written as n(10^0+10^1+10^2+........10^m)
  +can be written as n*(10^0+10^1+10^2+........10^m)
  for every (n,m) pair this relation gives a unique number. We can brute force every possible (n,m) pair
  till n(10^0+10^1+10^2+........10^m) exceeds the given number.
  contributed by Sameer Trivedi */
  