{"app":{"locale":"en-GB"},"problem":{"content":"

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 $1\\,000\\,000\\,007$.

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