It continues

Let $a_1, a_2, \ldots ,a_n$ be a sequence of posiitve integers except that $a_1\ge 0$ and $a_n \geq 2$ . Then the continued fraction $[a_1 ; a_2, \ldots , a_n ]$ is defined as $[a_1 ; a_2, \ldots , a_n ] = a_1 + \cfrac1{a_2 + \cfrac1{a_3 + \cfrac1{{\ddots}_{\quad + \cfrac1{a_n}}}}} .$ The following are two examples of how we can express a fraction in its continued fraction form: $\frac{12}5 = [2;2,2] = 2 + \frac1{2 + \frac12}\quad \text{ and }\quad \frac37 = [0;2,3] = 0 + \frac1{2 + \frac13},$ where it happens to be the case that three numbers $(2, 2, 2)$ or $(0, 2, 3)$ were used for each. Thus, if we define $f(A,B)$ as the total number of integers used in the continued fraction form of the fraction $\frac AB,$ where $A$ and $B$ are positive integers, then $f(12,5)= f(3,7) = 3.$

Find the maximum value of $f(m,2018)$ for all positive integers $m.$