Let $M$ denote a well-founded set-sized model of ZFC. The *descent value* of $M$ will be defined as the value of $n$ returned by the following process.

**Initialization.** Let $n$ equal $0$ and $X$ equal $M$.

**Step.** If $L_{\omega_1^X}^X$ doesn't satisfy ZFC according to $M$, halt and output $n$. Otherwise, increment $n$, let $X$ equal $L_{\omega_1^X}^X$, and repeat.

Question.Assuming sufficiently powerful large cardinal axioms, is it true that for every natural $n\geq 0,$ there exists a well-founded model $M$ of ZFC whose descent value is $n$?