For $\omega=e^{2\pi i/q}$ and $c_0,\ldots,c_{q-1}\in \mathbb Z$, $\sum\limits_{k=0}^{q-1}c_k\omega^k \equiv \sum\limits_{k=0}^{q-1}c_k\pmod p$.

Question

Let $q$ be a power of a prime $p$, and $\omega=e^{2\pi i/q}$. Suppose $c_0,\ldots,c_{q-1}\in \mathbb Z$ and that $\sum\limits_{k=0}^{q-1}c_k\omega^k \in \mathbb Z$. Then I have to show that $$\sum\limits_{k=0}^{q-1}c_k\omega^k \equiv \sum\limits_{k=0}^{q-1}c_k\pmod p.$$

Clearly $\omega$ is a primitive $q$th root of unity, where $q=p^r$ for some $r \in \mathbb N$. So $\omega$ has degree $\phi(q)=q-p^{r-1}$. Thus $\{1,\omega,\ldots,\omega^{q-p^{r-1}-1}\}$ is linearly independent over $\mathbb Q$. How can I proceed further? Need some help. Thanks.