Claim: $K_4 \not\subset G(V,E)$ Proof: Let $K_4 \subset G(V,E)\implies$ any edge of $K_4$ belongs at least four cycles.Contradiction. That is, any $4$ vertices belonging to the graph $G(V,E)$ can have at most $5$ edges among themselves.Let any $v_1,v_2,v_3,v_4 \in G(v)$ such that $(v_3,v_4)\not\in G(E)$ and and all remaining pairs are connected by edges.Let's say there is a cycle containing the vertices $(v_3,v_4)$. Then this cycle creates another cycle containing any $5$ edges we took at the beginning, so one of these $5$ edges belongs to at least $4$ cycles, which means there cannot be a new cycle containing the vertices $v_3,v_4$. Since the remaining edges are symmetric, let's take the edge $(v_2,v_4)$ without losing generality.Let's claim that the edge $(v_2,v_4)$ also belongs to another cycle. Let $v_2a_1a_2....a_nv_4$ be the cycle.In this case, any two of the vertices in the set $\{a_1,a_2,...a_n\}$ cannot form another cycle, otherwise there is an edge in the graph that belongs to at least $4$ cycles.Let $\{a,b,c\}$ our colors.When we match as $(v_1,a),(v_2,b),(v_3,c),(v_4,c)$, if $n$ is odd, $a_i$ is painted with color $a$ for $i\equiv 1 (mod 2)$ and the rest is $b$. If $n$ is even, the opposite coloring is done. (Additionally, case analysis can be performed for other subgraphs that can be created by deleting the edges of $K_4$. )