A8693. Graph Game

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):

1. Chose a node $x$ (it's common to chose weight-center) in tree $t$ . Let's call this step "Line A".
2. Deal with all paths that pass $x$ .
3. Then delete $x$ from tree $t$ .
4. After that $t$ becomes some subtrees.
5. Apply $solve$ on each subtree.

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):

1. $totalCost=totalCost+(size of t)$ . The operation "=" means assignment. $(Size of t)$ means the number of nodes in $t$ .
2. Choose a node $x$ in graph $t$ at random (uniformly among all nodes of $t$ ).
3. Then delete $x$ from graph $t$ .
4. After that $t$ becomes some connected components.
5. Apply $solve$ on each component.

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 $n$ ( $3<=n<=3000$ ) — the number of nodes and edges in the graph. Each of the next $n$ lines contains two space-separated integers $a_{i},b_{i}$ $(0<=a_{i},b_{i}<=n-1)$ indicating an edge between nodes $a_{i}$ and $b_{i}$ .

Consider that the graph nodes are numbered from $0$ to $(n-1)$ . 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 $10^{-6}$ .

Consider the second example. No matter what we choose first, the $totalCost$ will always be $3+2+1=6$ .