$C$. But then $c_1,c_2,\ldots, c_i, v$ is a cycle of length at least $\ell$ and at most $|C|-1$, contradicting the minimality of $C$. We conclude that every vertex of $V(G)\backslash V(C)$ has at most $\ell$ neighbours in $V(C)$. Hence the minimum degree of $G-V(C)$ is at least $(k-1)\ell-1$ and so, by induction, $G-V(C)$ contains $k-1$ vertex-disjoint cycles of length at least $\ell$. Together with $C$, these cycles yield the desired subgraph of $G$.

We remark that by the main result of [@CLNWY17], in fact the conclusion of Lemma \[lem:disjointlongcycles\] also holds for every graph with *average* degree larger than $k(\ell+1)-2$, which is sharp.\
The following auxiliary lemma will be applied in the main theorem to show that certain subgraphs of a minimum counterexample contain only short cycles.

\[lem:diameter\] Let $\ell$ and $g \geq 2\ell-1$ be positive integers. Let $G$ be a multigraph with diameter $<g/2$, such that every cycle of $G$ has either length $< \ell$ or length $>g$. Then every cycle of $G$ in fact has length $< \ell$.

For a contradiction, assume that $G$ has a cycle of length at least $\ell$, and let $C$ denote such a cycle of minimum length. Note that $|C| > g$. Consider 