How to conclude that $ \bar r \equiv 0$ (mod $m$)?

Question

Let $A$ be complete local ring with unique maximal ideal $m$.

Consider the equation $x^h-r=q(\gamma x^h)$, where $\gamma$ is a unit in $ A[[x]]$, $ r \in A[x]$ is a polynomial and $q \in A[[x]]$.

We have to show that $ r \in m[x]$.

So reducing the equation over $m$, we have

$x^h-\bar r=\bar q(\gamma x^h) \Rightarrow x^h+\gamma \bar qx^h=\bar r.$

How to conclude that $ \bar r \equiv 0$ (mod $m$)?