The prime factorization of number ~30~ is ~2 \times 3 \times 5~. Therefore, ~N~ is divisible by ~30~ if and only if it is divisible by ~2~, ~3~ and ~5~.

We know that a number is divisible by ~2~ if its last digit is even and it's divisible by ~3~ when the sum of its digits is divisible by ~3~ and, naturally, we know that it's divisible by ~5~ when its last digit is either ~0~ or ~5~. By combining these conditions, we conclude that ~N~ is divisible by ~30~ if and only if its last digit is ~0~ and the sum of its digits is divisible by ~3~. Therefore, if the number from the input data doesn't contain the digit ~0~ or the sum of its digits is not divisible by ~3~, we output -1.

What is obvious is that it is sufficient to sort the digits of number ~N~ in descending order to get the required number.

Implementations with time complexity of ~\mathcal O(n')~ or ~\mathcal O(n' \log n')~, where ~n'~ is the number of digits of ~N~, were sufficient to score all points on this task.