A question about irreducible characters of defect one

Question

Let $G$ be a finite group and $p$ a prime divisor of $|G|$. Also let $\mathrm{Irr}(G)$ and $\mathrm{IBr}(G)$ denote the set of irreducible complex characters and irreducible $p$-Brauer characters of $G$, respectively. It is well known that if $\chi \in \mathrm{Irr}(G)$, then the restriction of $\chi$ to $p$-regular elements of $G$ is a $p$-Brauer character, which is denoted by $\chi^0$. Therefore, we have $\chi^0=\Sigma_{\phi \in\mathrm{IBr}(G)}{d_{\chi\phi}~\phi}$. The integers $d_{\chi\phi}$ are called decomposition numbers.

Now assume that $\chi \in \mathrm{Irr}(G)$ is of defect one, that is $\chi(1)_p=\frac{|G|_p}{p}$. (For a natural number $n$, $n_p$ denotes the $p$-part of $n$ for the prime divisor $p$ of $n$). I heard in a recent talk that in this case, all the decomposition numbers $d_{\chi\phi}$ are less than or equal to $1$, and $\chi^0$ has at most $p-1$ irreducible constituents. I found that the first part of the assertion has been proved by Brauer. But still I don't see why the number of irreducible constituents of $\chi^0$ don't exceed $p$ in this case?

I would be grateful if you could explain it and enlighten my mind. Thank you in advance.