Let's assume that a Paf starts with some edge (v, u). Then, we can add to it any number of edges which are incident to vertex u. Assume that the last of them is an edge (u, w). Then, we can add any number of edges which are incident to vertex w. And so on.
So, for each paf there are two distinct vertices s and t. Let's denote the path between them as (v_1, v_2, v_3, ..., v_k), where s=v_1 and t=v_k. In this Paf edges (v_1, v_2), (v_2, v_3), ... , (v_{k-1}, v_k) will appear in this order, but we can insert more edges in between them. Let's assume that vertex v_2 has degree d. It means, that between edges (v_1, v_2) and (v_2, v_3) we can insert any partial permutation of d-2 edges which are incident to v_2.
How many such partial permutations are there? \sum_{i=0}^{d-2} \frac{(d-2)!}{i!}, simple combinatorics. We can mark this value as f(v). So, for every simple path consisting of at least one edge, we need to sum the product of values of f() for each vertex except for the endpoints, which is a straightforward tree dp.
First Solve: /boccadellaverita