All Covering spaces of the Annulus

Consider the annulus $$A:={(x,y)inmathbb{R}^{2}mid 1leq x^{2}+y^{2}leq 4}$$
I have to find all (connected) covering up to isomorphism.

My proof so far:

First of all, we have a homeomorphism $Acong S^{1}times [1,2]$. Therefore, we find $pi_{1}(A)cong pi_{1}(S^{1}times [1,2])cong pi_{1}(S^{1})=mathbb{Z}$. All the subgroups of $mathbb{Z}$ are of the form $mathbb{Z}n$ for $ninmathbb{N}$ and since $mathbb{Z}$ is abelian, all of them define distinct conjugacy classes of subgroups.

Now there is a theorem, which says that,if $A$ admits an universl cover, the then there are for each $n$ a covering $(widetilde{A},p)$ such that $p_{ast}(pi_{1}(widetilde{A}))cong nmathbb{Z}$ and this are all coverings up to isomorphism.

Now, I know that for $S^{1}$, we can define the maps $f_{n}:zmapsto z^{n}$, which define covering maps of $S^{1}$, such that $(f_{n})_{ast}(pi_{1}(S^{1}))cong nmathbb{Z}$. So therefore, my idea was to define the maps $g_{n}:S^{1}times [1,2]to Acong S^{1}times [1,2]$ exactly by $g_{n}(z,lambda):=(f_{n}(z),lambda)$. This should then be a covering with the claimed property $(g_{n})_{ast}(pi_{1}(S^{1}times [1,2]))cong nmathbb{Z}$.

Now to my question: Does this look right so far? and furthermore, the annulus $A$ is not simply connected, therefore all of my constructed coverings above are not universal. So I still have to find a universal cover, in order to show that I can apply the theorem above.