Let $a\neq -1,0,1$ be an integer. Write $a=(b^2c)^k$, where $b^2c$ is not a perfect power, and $c$ is squarefree. Artin's conjecture on primitive roots states that the asymptotic density of the set of prime numbers $p$ such that $a$ is a primitive root modulo $p$ is $$ \tag{$\star$} A(a)=\begin{cases} \left(\displaystyle\prod_{q|k,q\text{ prime}}\dfrac{q(q-2)}{q^2-q-1}\right)C,&c\not\equiv 1\pmod 4;\\ \left(\displaystyle\prod_{q|k,q\text{ prime}}\dfrac{q(q-2)}{q^2-q-1}\right)\left(1-\displaystyle\prod_{p|c,p\text{ prime}}\dfrac{-1}{p^2-p-1}\right)C,&c\equiv 1\pmod 4. \end{cases} $$ The formula is copied from here.
Everything looks well, except that at certain inputs the formula can't work: Let $q\neq 2$ be a prime and $m$ be a positive odd integer. Write $qm=q^rm'$ with $m'$ not divisible by $q$, and $r\in\mathbb{N}^*$, then we have $$ \tilde{q}^{qm}\text{ is a primitive root modulo }p\Leftrightarrow \tilde{q}^{m'}\text{ is a primitive root modulo }p, $$ where $$ \tilde{q}:=(-1)^{\frac{q-1}{2}}q=\begin{cases}q,&q\equiv 1\pmod 4;\\ -q,&q\equiv 3\pmod 4.\end{cases} $$ Here "$\Rightarrow$" is obvious; for "$\Leftarrow$", note that $\tilde{q}^{m'}\equiv 1\pmod 4$, so $\tilde{q}^{m'}$ being a quadratic nonresidue modulo $p$ implies that $p\not\equiv 1\pmod q$, then $$ \operatorname{ord}_{p}(\tilde{q}^{qm}) = \dfrac{\operatorname{ord}_{p}(\tilde{q}^{m'})}{\gcd(\operatorname{ord}_{p}(\tilde{q}^{m'}),q^r)} = \dfrac{p-1}{\gcd(p-1,q^r)} = p-1. $$ (As concrete examples, we have \begin{align*} &q=3,m=1:-27=(-3)^3\text{ is a primitive root modulo }p\Leftrightarrow -3\text{ is a primitive root modulo }p;\\ &q=5,m=1:3125=5^5\text{ is a primitive root modulo }p\Leftrightarrow 5\text{ is a primitive root modulo }p. \end{align*} However, the formula ($\star$) gives $$ \dfrac{A(\tilde{q}^{qm})}{A(\tilde{q}^{m'})} = \dfrac{q(q-2)}{q^2-q-1}\neq 1. $$ So my question is: Are numbers of the form $$ ((-1)^{\frac{q-1}{2}}q)^{qm},\quad q\neq 2\text{ prime},\quad m\text{ odd} $$ the only inputs for which the formula ($\star$) is false? (Assuming that Artin's conjecture is true, of course.) If not, could you please give a modified formula that would work for all $a$? Any hint/reference appreciated.