A frame that is not completely distributive

Question

Let $L$ be a frame, i.e., a complete lattice satisfying the infinite distributivity law $x\wedge \bigvee S=\bigvee \{x\wedge s\mid s\in S\}$. The totally below relation $x\lll y$ in $L$ means that if $y\le \bigvee S$, then $x\le s$ for some $s\in S$. It is well know that if $y= \bigvee \{x\in L\mid x\lll y\}$, then $L$ is completely distributive. An example of a frame that is not completely distributive is the lattice of regular open sets in $[0,1]$. However, in this lattice it is possible that $x\vee y=\top$ even if $x,y<\top$. What is an example of a frame that fails to be completely distributive due to $\bigvee \{x\in L \mid x\lll \top\}\ne \top$, in which $x\vee y=\top$ implies $x=\top$ or $y=\top$?