The graph formed by the towns and routes forms a tree structure. Upon rooting the tree at an arbitrary node (such as node ~1~), this problem can be solved with dynamic programming. For each node ~i~, let ~Up[i]~ be the number of different non-decreasing paths contained within ~i~'s subtree and ending at ~i~, and similarly let ~Down[i]~ be the number of different non-increasing paths contained within ~i~'s subtree and ending at ~i~. Paths of length ~0~ (spanning only ~1~ node) are included.

Let's consider how to compute these ~Up~ and ~Down~ values for each node ~i~. Firstly, we can set ~Up[i] = Down[i] = 1~ to include the path starting and ending at ~i~ (which is both non-decreasing and non-increasing). Next, consider each of ~i~'s children ~c~ which paths might be coming from. If ~D[i] > D[c]~, then all paths coming from ~c~ into ~i~ must non-decreasing, meaning that we can increment ~Up[i]~ by ~Up[c]~. Similarly, if ~D[i] < D[c]~, then we can increment ~Down[i]~ by ~Down[c]~. Finally, if ~D[i] = D[c]~, then both types of paths may occur, so we should both increment ~Up[i]~ by ~Up[c]~, and increment ~Down[i]~ by ~Down[c]~. Note that node ~i~'s ~Up~ and ~Down~ values only depend on those of its children, meaning that we can perform these computations recursively and populate these arrays for all nodes in ~\mathcal O(N)~ time.

What remains is computing the actual answer based on these ~Up~ and ~Down~ values. For each node ~i~, let's consider how many paths are contained within ~i~'s subtree and include ~i~ itself (in other words, paths which have ~i~ as their "topmost" node). If we can determine this value for every node, and add these ~N~ values together for all nodes, then every valid path will be counted exactly once. For the most part, this value for node ~i~ is simply ~Up[i] \times Down[i]~. However, some invalid paths also end up being counted in this fashion, which must then be subtracted. One such path is the path which starts and ends at ~i~, as the starting town is required to be different than the destination town, so we should subtract ~1~ for that. Aside from that, for each child ~c~ such that ~D[i] = D[c]~, this approach will include paths which come up from ~c~ to ~i~, and then back down to ~c~, which are also disallowed. As such, for each such child, we should simply subtract ~Up[c] \times Down[c]~ from the answer. These computations to tally up the total answer can be done during the dynamic programming process in an additional ~\mathcal O(N)~ time.