Resolution of $(f=0)$ where $f(x,y,z)=x^ay+y^az+\omega z^ax$

Question

Let $u\geq 2$ be an integer, $a=3u-1$, $\omega=e^{2\pi i /3u}$, and let $f$ be the polynomial $f(x,y,z)=x^ay+y^az+\omega z^ax$.

In Example 23 of https://www.intlpress.com/site/pub/files/_fulltext/journals/pamq/2008/0004/0002/PAMQ-2008-0004-0002-a001.pdf (p.217), it is written that: The singularity $(f = 0)$ can be resolved by $1$ blow up. The exceptional curve $C$ is smooth of genus $\binom{a}{2}$.

I am trying to understand and prove this statetment.

1. Does this mean the following: If $g:S'\to S=\{f=0\}$ is the minimal resolution of the isolated singularity at the origin of $S$ then $g^{-1}(0)$ is a curve of genus $\binom{a}{2}$?

2. If 1 is right, how can we show this? (A reference request to look up is enough. Actually I only know the algorithm for resolving an isolated surface singularity when the singularity is "simple", for example rational double points)