$2^{\omega(n)}$

Let $\omega(n)$ denote the number of distinct prime divisors of a positive integer $n$.
So $\omega(1) = 0$ and $\omega(360) = \omega(2^3\times 3^2 \times 5) = 3$.

Let $S(n)$ be $\sum_{d|n} 2^{\omega(d)}$.
E.g. $S(6) = 2^{\omega(1)} + 2^{\omega(2)} + 2^{\omega(3)} + 2^{\omega(6)} = 2^0 + 2^1 + 2^1 + 2^2 = 9$.

Let $F(n) = \sum_{i=2}^n S(i!)$. $F(10) = 4821$ 

Find $F(10\ 000\ 000)$. Give your answer modulo $1\ 000\ 000\ 087$.