## Bounded Divisors

### Problem 646 Published on Sunday, 9th December 2018, 04:00 am; Solved by 155

Let $n$ be a natural number and $p_1^{\alpha_1}\cdot p_2^{\alpha_2}...p_k^{\alpha_k}$ its prime factorisation.
Define the Liouville function $\lambda(n)$ as $\lambda(n) = (-1)^{\sum\limits_{i=1}^{k}\alpha_i}$.
(i.e. $-1$ if the sum of the exponents $\alpha_i$ is odd and $1$ if the sum of the exponents is even. )
Let $S(n,L,H)$ be the sum $\lambda(d) \cdot d$ over all divisors $d$ of $n$ for which $L \leq d \leq H$.

You are given:
$\bullet\, S(10! , 100, 1000) = 1457$
$\bullet\, S(15!, 10^3, 10^5) = -107974$
$\bullet\, S(30!,10^8, 10^{12}) = 9766732243224$.

Find $S(70!,10^{20}, 10^{60})$ and give your answer modulo $1\,000\,000\,007$.