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|?

\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.