2013 AMC 10B Problem 20

The number $2013$ is expressed in the form $$2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!},$$ where $a_1\ge a_2\ge\cdots\ge a_m$ and $b_1\ge b_2\ge\cdots\ge b_n$ are positive integers and $a_1+b_1$ is as small as possible. What is $|a_1-b_1|$ ?