$\sum^{2019}_{n=1}a_n^2\bmod 8$

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Let $a_1=20$ , $a_2=19$ , and $a_n=(a_{n-1}+a_{n-2}) \bmod 987563124$ for $n>2$ .

Find the remainder when $\displaystyle \sum^{2019}_{n=1}a_n^2$ is divided by $8$ .