Fibonacci and 2018

Let $(F_{n})_{n \ge 1}$ be Fibonacci's sequence defined by the following recurrence relation: $F_1 = 1, \space F_2 = 1$ and $F_n = F_{n - 1} + F_{n - 2}$ for all natural number $n \ge 3$ .

Then, $\displaystyle \large \sum_{n = 1}^{\infty} \frac{F_n}{(2018^{2018})^n}$ can be written as $\frac{A}{B}$ being $A, B$ coprime positive integers.

What is the smallest positive integer $C$ such that $B \equiv C \text{ (mod 2018) }$ ?