The total number of prime factors of $n$, counted with multiplicity, is denoted $\mathrm{\Omega }(n)$.
For example, $\mathrm{\Omega }(12)=3$, 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 $\mathrm{\Omega }(d)$ is divisible by $m$.
For example, $D(24,3)=1+8+12=21$.

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

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

Find $D(1000★,1000)$. Give your answer modulo $999999001$.