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 $\mathcal{O}(n^{\prime })$ or $\mathcal{O}(n^{\prime }\log n^{\prime })$, where $n^{\prime }$ is the number of digits of $N$, were sufficient to score all points on this task.