Recall, a group $G$ is an almost simple group if there is an non-abelian simple group $S$ such that $S \leq G \leq \operatorname{Aut}(S)$
We will define a $k$-almost simple as follows.
When $0$--almost simple is a non-abelian simple group.
For $k \geq 1$, a group $G$ is $k$-almost simple if there exists a group $S$ such that S is $k-1$ almost simple and $S \leq G \leq \operatorname{Aut}(S)$.
Notice that under this definition, $1$-almost simple groups are just the almost simple groups.
Question: For any finite group $G$, does there exist a $k>0$ such that $G$ is contained in some finite $k$-almost simple group?