The Chinese Remainder Theorem · Superchallenge problem

For how many integers $a$ with $0 \le a < 1001$ does the congruence $x^2 \equiv a \pmod{1001}$ have exactly $4$ solutions $x$ with $0 \le x < 1001$? (Here $1001 = 7 \cdot 11 \cdot 13$.)