How many homeomorphically distinct graphs exist such that they contain a path that is both Eulerian and Hamiltonian?

**Definitions:**

- A
**walk**on a graph is a sequence \((v_0, e_1, v_1, e_2, v_2, \ldots, e_k, v_k)\), where \(v_0, v_1, \ldots, v_k\) are vertices of the graph, \(e_1, e_2, \ldots, e_k\) are edges of the graph, and \(e_i = \{v_{i-1}, v_i\}\) for all \(i = 1, 2, \ldots, k\). - An
**Eulerian path**is a walk that uses each edge exactly once. That is, \(e_1, e_2, \ldots, e_k\) are all distinct, and they are exactly the edges of the graph in some order. - A
**Hamiltonian path**is a walk that uses each vertex exactly once. That is, \(v_0, v_1, \ldots, v_k\) are all distinct, and they are exactly the vertices of the graph in some order. - A
**subdivision**of an edge \(e = \{u,v\}\) in a graph gives a modified version of the graph, having an additional vertex \(w\) and two additional edges \(\{u,w\}, \{v,w\}\), and missing the original edge \(e\). A subdivision of a graph is the result of subdividing any number of the edges of the graph. (An edge that's the result of some earlier subdivision may still be subdivided.) - Two graphs \(G, G'\) are
**homeomorphic**if there is some graph \(H\) that is both isomorphic to a subdivision of \(G\) and isomorphic to a subdivision of \(G'\). Informally, if \(G\) and \(G'\) are drawn with vertices as invisible points and edges as arcs, then two homeomorphic graphs "look the same." Two graphs are**homeomorphically distinct**if they are not homeomorphic.

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