$f(v_i) = \lfloor \frac{q+1}{2} \rfloor$ for all $i \in \{r+1,\ldots,q\}$. We obtain $f(N(u_i)) = f(Q) - f(u_i) + f(v_i) = f(Q) - r + i$ for all $i \in [q]$. Then, for $j < k$, we have $f(N(u_j)) < f(N(u_k))$. Regarding the edge $(u_i, v_i)$, we first analyze when $i = 1$. Note that $f(u_1) = r$, $f(u_2) = r - 1$ and $f(u_q) = 1$, then $f(N(u_1)) = f(Q) - r + 1 \geq f(u_1) + f(u_2) + f(u_q) - r + 1 = r + 1 > r = f(N(v_1))$. If $i \geq 2$, $f(N(u_i)) > f(N(u_1)) > f(N(v_1)) = r \geq f(u_i) = f(N(v_i))$.

Now, we consider that $G$ is thick. If $q = 2$ then $G$ is isomorphic to a thin headless spider of order 2. Hence, assume that $q \geq 3$. Consider the assignment $f:V \rightarrow [r]$ such that $f(u_i) = i$ and $f(v_i) = 1$ for all $i \in [r]$, and $f(u_i) = r$ and $f(v_i) = i-r+1$ for all $i \in \{r+1,\ldots,q\}$. We obtain $f(N(u_i)) = f(V) - f(u_i) - f(v_i) = f(V) - i - 1$ for all $i \in [q]$. Then, for $j < k$, we have $f(N(u_j)) > f(N(u_k))$. Regarding 