Eccoci.

The total number of prime factors of $n$, counted with multiplicity, is denoted $\Omega(n)$.
For example, $\Omega(12)=3$, counting the factor $2$ twice, and the factor $3$ once.

Define $D(n, m)$ to be the sum of all divisors $d$ of $n$ where $\Omega(d)$ is divisible by $m$.
For example, $D(24, 3)=1+8+12=21$.

The superfactorial of $n$, often written as $n\$$, is defined as the product of the first $n$ factorials: $$n\$=1!\times 2! \times\cdots\times n!$$ The superduperfactorial of $n$, we write as $n\bigstar$, is defined as the product of the first $n$ superfactorials: $$n\bigstar=1\$ \times 2\$ \times\cdots\times n\$ $$

You are given $D(6\bigstar, 6)=6368195719791280$.

Find $D(1\,000\bigstar, 1\,000)$. Give your answer modulo $999\,999\,001$.