Problem:
Score : 600 points

Problem Statement

Given is a tree T with N vertices. The i-th edge connects Vertex A_i and B_i (1 \leq A_i,B_i \leq N).

Now, each vertex is painted black with probability 1/2 and white with probability 1/2, which is chosen independently from other vertices. Then, let S be the smallest subtree (connected subgraph) of T containing all the vertices painted black. (If no vertex is painted black, S is the empty graph.)

Let the holeyness of S be the number of white vertices contained in S. Find the expected holeyness of S.

Since the answer is a rational number, we ask you to print it \bmod 10^9+7, as described in Notes.

Notes

When you print a rational number, first write it as a fraction \frac{y}{x}, where x, y are integers, and x is not divisible by 10^9 + 7
(under the constraints of the problem, such representation is always possible).

Then, you need to print the only integer z between 0 and 10^9 + 6, inclusive, that satisfies xz \equiv y \pmod{10^9 + 7}.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i,B_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the expected holeyness of S, \bmod 10^9+7.

Sample Input 1

3
1 2
2 3

Sample Output 1

125000001

If the vertices 1, 2, 3 are painted black, white, black, respectively, the holeyness of S is 1.

Otherwise, the holeyness is 0, so the expected holeyness is 1/8.

Since 8 \times 125000001 \equiv 1 \pmod{10^9+7}, we should print 125000001.

Sample Input 2

4
1 2
2 3
3 4

Sample Output 2

375000003

The expected holeyness is 3/8.

Since 8 \times 375000003 \equiv 3 \pmod{10^9+7}, we should print 375000003.

Sample Input 3

4
1 2
1 3
1 4

Sample Output 3

250000002

The expected holeyness is 1/4.

Sample Input 4

7
4 7
3 1
2 6
5 2
7 1
2 7

Sample Output 4

570312505

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    Node[] nodes = new Node[N];
    for (int i = 0; i < N; i++) nodes[i] = new Node();

    for (int i = 0; i < N - 1; i++) {
      int a = scanner.nextInt() - 1;
      int b = scanner.nextInt() - 1;
      nodes[a].adjacent.add(nodes[b]);
      nodes[b].adjacent.add(nodes[a]);
    }
    dfs(nodes[0], null);

    int M = 1_000_000_007;
    long res = N * modpow(2, M - 2, M) % M;
    for (Node node : nodes) {
      res += (node.adjacent.size() - 1) * modinv(modpow(2, N, M), M) % M;
      if (res >= M) res -= M;
      for (Node c : node.children) {
        res -= modinv(modpow(2, N - c.size, M), M);
        if (res < 0) res += M;
      }
      if (node.parent != null) {
        res -= modinv(modpow(2, node.size, M), M);
        if (res < 0) res += M;
      }
    }
    System.out.println(res);
  }

  private static long modinv(long a, long mod) {
    return modpow(a, mod - 2, mod);
  }

  private static long modpow(long a, long n, long mod) {
    long res = 1;
    while (n > 0) {
      if (n % 2 == 1) res = (res * a) % mod;
      a = (a * a) % mod;
      n >>= 1;
    }
    return res % mod;
  }

  private static int dfs(Node node, Node parent) {
    node.parent = parent;
    node.size = 1;
    for (Node a : node.adjacent) {
      if (a != parent) {
        node.children.add(a);
        node.size += dfs(a, node);
      }
    }
    return node.size;
  }

  private static class Node {
    private Node parent;
    private int size;
    private List<Node> children = new ArrayList<>();
    private List<Node> adjacent = new ArrayList<>();
  }
}