I am interested in bounding the following Salie-type ("twisted Kloosterman") sum

$$
S(a,b,\beta) = \sum\_{x \in \mathbb{Z}/{p^{\beta}}\mathbb{Z}} \left( \frac{x}{p^{\beta}} \right) \chi(ax + bx^{-1}).
$$

Here, $\left( \frac{\cdot}{q} \right)$ denotes the Jacobi symbol, $\chi(x) = \exp(2 \pi i x /p^{\beta})$, $p$ is an odd prime, $\beta$ is a positive integer, and I am always assuming $\gcd(p,ab) = 1$.

I am trying to find a bound for these Salie-type sums. When $\beta = 1$, the bound $|S(a,b,1)| \leq 2 \sqrt{p}$ is due to Salie. When $\beta \geq 2$ is even, the Jacobi symbol is identically $1$ and so the sum reduces to a Kloosterman sum

$$
K(a,b,\beta) = \sum\_{x \in U(\mathbb{Z}/{p^{\beta}} \mathbb{Z})} \chi(ax + bx^{-1})
$$

and the well known Weil bound applies giving $|K(a,b,\beta)| \leq (\beta + 1) p^{\beta/2}$. Here, $U(R)$ denote the set of units in $R$.

My question is: **How does one bound the sum $S(a,b,\beta)$ in the case $\beta \geq 3$ is odd?** It seems reasonable that the bound $|S(a,b,\beta)| \leq (\beta + 1) p^{\beta/2}$ should apply here as well.

Thank you for your help.