\ge0.$  If $n=0$ there are two possibilities: $r=0$ and $r=0.$ Let $r=0$ then $s=0$ and by [@fpr0 Proposition 0.00] we obtain $(q^n-0)^0n^0<\frac{0}{0}q^{0\cdot0}<|\overline{x}^{\overline{G}}| \le |x^G|$ for $q>0$, case $q=0$ can be checked by GAP. If $r=p=0$ then $s=0$ and $|x^G|>\frac{0}{0}q^{0\cdot0}.$ 