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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
  "bufio"
  "os"
  "strconv"
  "strings"
  "fmt"
)

var reader = bufio.NewReaderSize(os.Stdin, 1000000)
var writer = bufio.NewWriter(os.Stdout)
func NextLine() string {
  buffer := make([]byte, 0)
  for true {
    line, isPrefix, err := reader.ReadLine()
    if err != nil { panic(err) }
    buffer = append(buffer, line...)
    if !isPrefix { break }
  }
  return string(buffer)
}
func NextInt() int {
  n, _ := strconv.Atoi(NextLine())
  return n
}
func NextIntVec() []int {
  L := strings.Split(NextLine(), " ")
  N := make([]int, len(L))
  for i := range N {
    N[i], _ = strconv.Atoi(L[i])
  }
  return N
}
func Write(s interface{}) {
  fmt.Fprintln(writer, s)
}
func Output() {
  _ = writer.Flush()
}

func Sum(A []int) int {
  sum := 0
  for _, a := range A {
    sum += a
  }
  return sum
}

const p = 1000000007
const p2 = 500000004

func main() {
  N := NextInt()
  T := make([][]int, N)
  for i := 1; i < N; i++ {
    AB := NextIntVec()
    A, B := AB[0] - 1, AB[1] - 1
    T[A] = append(T[A], B)
    T[B] = append(T[B], A)
  }
  P := make([]int, N)
  for i := range P {
    if 0 < i { P[i] = -1 }
  }
  Q := make([]int, N)
  pos := 1
  for _, q := range Q {
    for _, adj := range T[q] {
      if 0 <= P[adj] { continue }
      P[adj] = q
      Q[pos] = adj
      pos++
    }
  }
  ST := make([][]int, N)
  for i := range ST {
    ST[i] = make([]int, 0)
  }
  for i := N - 1; 0 < i; i-- {
    q := Q[i]
    s := Sum(ST[q]) + 1
    ST[P[q]] = append(ST[P[q]], s)
    ST[q] = append(ST[q], N - s)
  }
  B := make([]int, N)
  B[0] = 1
  for i := 1; i < N; i++ {
    B[i] = B[i - 1] * p2 % p
  }
  ans := 0
  for _, S := range ST {
    R := make([]int, N)
    for i, s := range S {
      R[i] = B[N - 1 - s] * (p + 1 - B[s]) % p
    }
    cnt := 1 - B[N - 1] - Sum(R)
    if cnt < 0 { cnt += (-cnt / p) * p }
    if cnt < 0 { cnt += p }
    ans += p2 * cnt % p
    ans %= p
  }
  Write(ans)
  Output()
}