Let $R$ be a Cohen-Macaulay ring of dimension $1$ with total ring of fractions $Q(R)$. Let $R\subseteq S \subseteq Q(R)$ be a ring such that $Q(S)=Q(R)$ and $S$ is a (module) finite extension of $R$. Hence, $S$ is a Noetherian ring of dimension $1$. Is $S$ also Cohen-Macaulay?
I can see that this is true if $R$ is reduced, because then $S$ is also reduced, and reduced rings of dimension $1$ are Cohen-Macaulay. But I am not sure what happens otherwise.
Please help.