Bulgarian Spring Tournament

We reason inductively on $n$, not writing the number $1$ at any vertex. For $n=1$ the requested is clear, for $n=2$ an example with $e=1$ is $(2,4)$ and an example with $e=0$ is $(2,3)$. For $n=3$ example with $e=0$ is $3,5,7$, example with $e=1$ is $2,4,7$, example with $e=2$ is $2,4,10$, an example with $e=3$ is $2,4,8$. For $n\ge 4$ we have $n-1\le \frac{(n-1)(n-2)}{2}$, so at least one of $e\ge n-1$ and $e\le \frac{(n-1)(n-2)}{2}$ is satisfied. Let $e\ge n-1$ first. In an example with $n-1$ vertices and $e-(n-1)$ edges, we add a vertex (of degree $n-1$) by writing in it a prime number $p$ greater than the numbers in the other vertices, then we multiply the numbers in the remaining vertices by $p$ – this does not spawn new edges between the remaining vertices. Now let $e\le \frac{(n-1)(n-2)}{2}$. In an example with $n-1$ vertices and $e$ edges, we add a vertex (of degree $0$), writing in it a prime number $p$ greater than the numbers in the other vertices, and do not change the numbers in the other vertices – it does not spawn new edges.