Let $f$ be an automorphic form on $GL_3$ for $\Gamma_0(p)$ with $p$ being a prime (see Bump or Goldfeld's books for definitions). Recall that, in this paper-"The Voronoi formula on GL(3) with ramification", Zhou has formulated that, for any integers $a,c$ satisfying $(a,c)=1$ with $p|c$, one roughly has the corresponding relation: $$\sum _{n\sim X}{A_f(n,1)e\left(\frac{an}{c}\right)} \omega \left ({n} \right) \leftrightsquigarrow c \sum_{\substack{m_2\neq 0}} \sum_{m_1|c} \frac{A_f(m_1,m_2)}{m_1|m_2|} S(a, m_2;c/m_1) \mho \left ( \frac{m^2_1|m_2|X}{c^3} ;\omega\right) .$$

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: $$ \sum_{n\sim X} A_{F}(n,1,1) e \left( \frac{an}{c} \right) \omega \left ({n} \right) \leftrightsquigarrow c\sum_{d_1|c}\sum_{d_2|\frac{c}{d_1}}\sum_{m\neq 0} \frac{\overline{A_F(m, d_2,d_1)}}{|m|d_2d_1} {KL}_2(\overline{a}, m,c;(1,1) , (d_1,d_2) )\, \mho^\prime \left ( \frac{mXd^2_2 d^3_1}{c^4} ;\omega\right),\;?? $$ where the hyper-Kloosterman sum is defined as $${KL}_2(n, m,c;(1,1), (d_1,d_2))= \sideset{_{}^{}} {^{\ast}_{}}\sum\limits_{x_1 \bmod \frac{q_1c}{d_1} } \,\, \sideset{_{}^{}}{^{\ast}_{}}\sum\limits_{x_2 \bmod \frac{c}{d_1d_2} } e{\left (\frac{d_1 x_1 n}{c}+\frac{d_2\overline{x_1x_2}}{\frac{c}{d_1}}+\frac{{x_2}m}{ \frac{c}{d_1d_2}}\right )} .$$ Note that when the level is trivial, the formula above for $GL_4$ is already known; see, for example, V. Chandee and 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 (MR4331312, Zbl 1494.11050). 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.