Let $p$ be an odd prime and consider the congruence subgroup of $\mathrm{SL}_{2}(\mathbb{Z})=\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \ \Big | \ a,b,c,d\in\mathbb{Z}, \ ad-bc=1 \right\}$,
$$ \Gamma_{1}(p)=\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in\mathrm{SL}_{2}(\mathbb{Z}) \ \Big | \ c\equiv 0\pmod p, \ a\equiv d\equiv1\pmod p\right\}.$$
I am trying to find the set of cusps of $\Gamma_{1}(p)$, i.e. a set of representatives for the $\Gamma_{1}(p)$-orbits in $\mathbb{P}^{1}(\mathbb{Q})=\mathbb{Q}\cup\{\infty\}$. (Here the action is defined as
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}t=\left\{\begin{matrix} \frac{at+b}{ct+b}, & t\in\mathbb{Q} \\ \frac{a}{c}, & t=\infty \end{matrix}\right.$$
where $\frac{at+b}{ct+b}:=\infty$, if $ct+b=0$.)
I know how to find the set of cusps of
$$\Gamma_{0}(p)=\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in\mathrm{SL}_{2}(\mathbb{Z}) \ \Big | \ c\equiv 0\pmod p\right\}=\left\{\begin{pmatrix} a & b \\ cp & d \end{pmatrix} \ \Big | a,b,c,d\in\mathbb{Z}, \ ad-bcp=1\right\},$$
as the orbit of $\infty$ is
\begin{align} \Gamma_{0}(p)\cdot\infty&=\left\{\begin{pmatrix} a & b \\ cp & d \end{pmatrix}\infty \ \Big | \ a,b,c,d\in\mathbb{Z}, \ ad-bcp=1, \ \right\} \\ &=\left\{\frac{a}{cp} \ \Big | \ a,c\in\mathbb{Z}, \ (a,cp)=1, \ \right\} \\ &=\left\{\frac{r}{s} \ \Big | \ r,s\in\mathbb{Z}, \ (r,s)=1, \ p|s \ \right\} \end{align}
and, likewise, the orbit of $0$ is
$$\Gamma_{0}(p)\cdot 0=\left\{\frac{r}{s} \ \Big | \ r,s\in\mathbb{Z}, \ (r,s)=1, \ p\not|s \ \right\}$$.
Hence, $\mathbb{P}_{1}(\mathbb{Q})=\Gamma_{0}(p)\cdot\infty\cup\Gamma_{0}(p)\cdot 0$.
Now, as $\Gamma_{1}(p)\subset\Gamma_{0}(p)$ we have that there will be more cusps than $0$ and $\infty$ but I do not know how to find the rest. Any help?