B'In computer science, there is a method called "Divide And Conquer By Node" to solve some hard problems about paths on a tree. Let 's desribe how this method works by function: solve(t) (t is a tree): This ends when t has only one node because after deleting it, there 's nothing. Now, WJMZBMR has mistakenly believed that it 's ok to chose any node in "Line A". So he 'll chose a node at random. To make the situation worse, he thinks a "tree" should have the same number of edges and nodes! So this procedure becomes like that. Let 's define the variable totalCost. Initially the value of totalCost equal to 0. So, solve(t) (now t is a graph): He 'll apply solve on a connected graph with n nodes and n edges. He thinks it will work quickly, but it 's very slow. So he wants to know the expectation of totalCost of this procedure. Can you help him? The first line contains an integer n (3 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89 xe2 x89 xa4 xe2 x80 x893000) -- the number of nodes and edges in the graph. Each of the next n lines contains two space-separated integers ai, xe2 x80 x89bi (0 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89ai, xe2 x80 x89bi xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89- xe2 x80 x891) indicating an edge between nodes ai and bi. Consider that the graph nodes are numbered from 0 to (n xe2 x80 x89- xe2 x80 x891). It 's guaranteed that there are no self-loops, no multiple edges in that graph. It 's guaranteed that the graph is connected. Print a single real number -- the expectation of totalCost. Your answer will be considered correct if its absolute or relative error does not exceed 10 xe2 x80 x89- xe2 x80 x896. Consider the second example. No matter what we choose first, the totalCost will always be 3 xe2 x80 x89+ xe2 x80 x892 xe2 x80 x89+ xe2 x80 x891 xe2 x80 x89= xe2 x80 x896.'...