// This is a simple iterative solution to a puzzle.  It illustrates Frink's
// "multifor" loop and permutation algorithms, and bailing out of loops with
// labeled targets.
//
// Puzzle definition:
// https://twitter.com/snegopa/status/980067722628337665

count = 0
x = new range[1, 9]

NUMS:
multifor nums = [x, x, x, x]
{
   for order = nums.lexicographicPermute[]
   {
      n = parseInt[join["", order]]
      if n mod 12 != 0
         next NUMS
   }

   count = count + 1
   println[join["", nums]]
}

println["Solution is $count"]