How to find $R$ such that $\prod_{p\mid q} \Big(p-2-r +O\left( \frac{1}{p-2}\right) \Big)=\prod_{p\mid q} (p-2-r) +O(R)$.

Question

suppose $q$ a square free integer (so that $\prod_{p\mid q} p =q$). Let $r$ be a positive integer, and assume that $q$ is not divisible by any prime less than or equal to $r+2$.

Now, suppose that I have $\prod_{p\mid q} \Big(p-2-r +O\left( \frac 1 {p-2}\right) \Big)$. I want to express this in terms in $\prod_{p|q} (p-2-r).$

i.e. I want to find $R$ such that $\prod_{p\mid q} \Big(p-2-r +O\left( \frac{1}{p-2}\right) \Big)=\prod_{p\mid q} (p-2-r) +O(R)$.

How do I find $R$?

Thanks