everyone. I have a question need some help from the experts in this website. Let $f$ be a an automorphic form on $GL_3$ for $\Gamma_0(p)$ with $p$ being a prime. Recall that, in this paper-"The Voronoi formula on GL(3) with ramification“ (https://arxiv.org/pdf/1806.10786), Zhou has formulated that, for any integers $a,c$ satisfying $(a,c)=1$ with $p|c$, we roughly have the corresponding relation:
$$\sum _{n\sim X}{A_f(n,1)e(\frac{an}{c})} \omega \left ({n} \right)    \leftrightsquigarrow  c \sum_{\substack{m_2\neq 0}} \sum_{m_1|c} 
 \frac{A_f(m_1,m_2)}{m_1|m_2|}  S(a, m_2;c/m_1) \mho \left ( \frac{m^2_1m_2X}{c^3}  ;\omega\right) .$$

Now, consider $F$ be an automorphic form on $GL_4$ of level $p$. My question is: in the case of $p|c$, whether or not we have a similar formula like:
$$\sum_{n\neq 0}  A_{F}(n,1,1) 
 e \left( \frac{an}{c} \right)  
 \omega \left ({n} \right) \leftrightsquigarrow  c\sum_{d_1|c}\sum_{d_2|\frac{c}{d_1}}\sum_{m\neq 0} \frac{\overline{A_F(m, d_2,d_1)}}{|m|d_2d_1} {KL}_2(\overline{a}, m,c;(1,1) , d_1,d_2 )\, \mho^\prime \left ( \frac{mXd^2_2 d^3_1}{c^4}  ;\omega\right)，??$$
where the hyper-Kloosterman sum is defined as 
$${KL}_2(n, m,c;(1,1),  (d_1,d_2))=\sum^\ast_{x_1 \bmod \frac{c}{d_1} } \sum^\ast_{x_2 \bmod \frac{c}{d_1d_2} } e{\left (\frac{d_1 x_1 n}{c}+\frac{d_2\overline{x_1x_2}}{\frac{c}{d_1}}+\frac{{x_2}m}{ \frac{c}{d_1d_2}}\right )} .$$
Note that when the level is trivial, the formula above for $GL_4$ is already known; see, for example, V. Chandee-X. Li's paper (The second moment of $GL(4) \times GL(2)$ $L$-functions at special points). Besides, in the special case where $(p,c)=1$, one has an analog for $GL_4$ Maass form, which has a similar structure like Eqn. (2) in Zhou's above mentioned paper; see, for example, Jeanine Van Order's paper (https://arxiv.org/pdf/1903.06690). I still can not be sure what looks like for the Voronoi formula on $GL(4)$ when $p|c$. 

If any expert here knows something on this question, please show a guide or certain references.

Many many thanks in advance.