Waldspurger formula for Fourier coefficients of forms in Kohnen's space.

Let $f\in S_{k+1/2}^+(4q)$ be a newform in Kohnen’s space for $q$ an odd, square-free integer. For simplicity, assume that $k$ is even. Let $$f(z) = \sum_{\substack{n\geq 1\\ n\equiv 0,1\mod 4}}a_f(n)e(nz)$$ denote the Fourier expansion of $f$ at the cusp $\infty$. Let $D>0$ be a fundamental discriminantwith $(D,q)=1$ and $\left(\frac{D}{p}\right) = \varepsilon_p$ for all $p|q$, where $\varepsilon_p$ are the Atkin-Lehner signs of $f$. By Corollary 1 in Kohnen's paper (https://link.springer.com/article/10.1007/BF01455989), we know that: $$ \frac{|a_f(D)|^2}{\langle f,f\rangle} = \frac {(k-1)!}{\pi^k}2^{\omega(q)}|D|^{k-1/2}\frac{L(1/2,F\otimes \chi_{D})}{\langle F,F\rangle},$$ where $F\in S_{2k}^{\text{new}}(q)$ is a Hecke normalised newform that corresponds to $f$ under the Shimura correspondence, $\omega(q)$ is the number of prime facotrs of $q$, and $\langle f,f\rangle$, $\langle F,F\rangle$ are the $L^2$-norms of $f,F$.

Now let $D>0$ be a fundamental discriminant with $d = (D,q)>1$. and let $S_d = \{p \text{ prime}: p|\frac{q}{d}\}$. My question is the following: Is it true that if $\left(\frac{D}{p}\right) = -\varepsilon_p$ for some $p\in S_d$ then we have $a_f(D)=0$, and if $\left(\frac{D}{p}\right) = \varepsilon_p$ for all $p\in S_d$ then $$\frac{|a_f(D)|^2}{\langle f,f\rangle} = \frac {(k-1)!}{\pi^k}2^{\omega(q)-\omega(d)}|D|^{k-1/2}\frac{L(1/2,F\otimes \chi_{D})}{\langle F,F\rangle} \prod_{p\in S_d}\frac{p}{p+1} $$?