Problem 956

Super Duper Sum

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 $999\ 999\ 001$.

超级无敌和

记$n$的质因数数目（包括重复）为$\Omega(n)$。

例如，$\Omega(12)=3$，包括两个质因数$2$和一个质因数$3$。

定义$D(n, m)$为$n$的所有满足$\Omega(d)$能被$m$整除的因数$d$之和。

例如，$D(24, 3)=1+8+12=21$。

$n$的超级阶乘，通常记作$n\$$，定义为前$n$个阶乘的乘积：
$$n\$=1!\times 2! \times\cdots\times n!$$
$n$的超级无敌阶乘，我们记作$n\bigstar$，定义为前$n$个超级阶乘的乘积：
$$n\bigstar=1\$ \times 2\$ \times\cdots\times n\$ $$

已知$D(6\bigstar, 6)=6368195719791280$。

求$D(1\ 000\bigstar, 1\ 000)$，并对$999\ 999\ 001$取余作为你的答案。