In a sequence of natural numbers $a_1$, $a_2$, $\dots$, $a_{1999}$, the expression $a_n - a_{n-1} - a_{n-2}$ is divisible by $100$ for $3 \leq n \leq 1999$. It is known that $a_1 = 19$ and $a_2 = 99$. Find the remainder when $a_1^2 + a_2^2 + \dots + a_{1999}^2$ is divided by $8$.

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