For every integer $a > 1$, an infinite list of integers is constructed $L(a)$, as follows:

1. $a$ is the first number in the list $L(a)$.
2. Given a number $b$ in $L(a)$, the next number in the list is $b + c$, where $c$ is the largest integer that divides $b$ and is smaller than $b$.

Find all the integers $a > 1$ such that $2002$ is in the list $L(a)$.

Let's think step by step and output the final answer within \boxed{}.
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