Verifying Hamiltonian Cycle solution in O(n^2), n is the length of the encoding of G

Question

In the textbook of CLRS, 'ch. 34.2 Polynomial-time verification' it says the following:

Suppose that a friend tells you that a given graph G is hamiltonian, and then offers to prove it by giving you the vertices in order along the hamiltonian cycle. It would certainly be easy enough to verify the proof: simply verify that the provided cycle is hamiltonian by checking whether it is a permutation of the vertices of $V$ and whether each of the consecutive edges along the cycle actually exists in the graph. You could certainly implement this verification algorithm to run in $O(n^2)$ time, where $n$ is the length of the encoding of $G$.

To me, for each consecutive pair $(u,v)$ of the given cycle, we could verify if it's an edge in $G$. Further we could use some color coding for each vertex to ensure we don't revisit a vertex. By doing so, we could verify if the given cycle is Hamiltonian in $O(E)=O(m^2)$ time where $m$ is the number of vertices in $G$. Further we can see the minimum encoding $n$ of $G$ is $m^2=n$. Thus $O(E)=O(m^2)=O(n)$. Can anyone help me understand, why it is mentioned as $O(n^2)$ instead!