CF629E Famil Door and Roads

Famil Door’s City map looks like a tree (undirected connected acyclic graph) so other people call it Treeland. There are $ n $ intersections in the city connected by $ n-1 $ bidirectional roads. There are $ m $ friends of Famil Door living in the city. The $ i $ -th friend lives at the intersection $ u_{i} $ and works at the intersection $ v_{i} $ . Everyone in the city is unhappy because there is exactly one simple path between their home and work. Famil Door plans to construct exactly one new road and he will randomly choose one among $ n·(n-1)/2 $ possibilities. Note, that he may even build a new road between two cities that are already connected by one. He knows, that each of his friends will become happy, if after Famil Door constructs a new road there is a path from this friend home to work and back that doesn't visit the same road twice. Formally, there is a simple cycle containing both $ u_{i} $ and $ v_{i} $ . Moreover, if the friend becomes happy, his pleasure is equal to the length of such path (it's easy to see that it's unique). For each of his friends Famil Door wants to know his expected pleasure, that is the expected length of the cycle containing both $ u_{i} $ and $ v_{i} $ if we consider only cases when such a cycle exists.

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Consider the second sample. 1. Both roads $ (1,2) $ and $ (2,3) $ work, so the expected length if  2. Roads $ (1,3) $ and $ (2,3) $ make the second friend happy. Same as for friend $ 1 $ the answer is $ 2.5 $ 3. The only way to make the third friend happy is to add road $ (2,3) $ , so the answer is $ 3 $