Generating set for submodule of free module over a PIR

Question

Let $R$ be a commutative principal ideal ring (not necessarily a domain), $S \cong R^n$ a free $R$-module. We know that any submodule $M$ of $S$ has $m := \operatorname{length}_R(M) \leq n \cdot \operatorname{length}_R(R)=\operatorname{length}_R(S)$. Is it true that there exists $\{u_i\}_{1\leq i \leq n}$ a basis of $S$ and scalars $d_1, \dots, d_m$ with $d_1 \mid d_2 \mid \dotsc \mid d_m$ such that $\{d_i u_i\}_{1\leq i \leq m}$ is a generating set for $M$? I think that this is true and an easy way to show it is simply by using the Smith normal form (which exists for general PIRs), is there something wrong with this reasoning? Is it true only when $R$ is a PID?