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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 6)

mod = 10 ** 9 + 7

N = int(input())
vec = [[] for _ in range(N)]
for _ in range(N - 1) :
  A, B = map(int, input().split())
  vec[A-1].append(B-1)
  vec[B-1].append(A-1)

sub_size = [[] for _ in range(N)]

def dfs(cur, pre) :
  s = 0
  for nex in vec[cur] :
    if nex == pre :
      continue
    t = dfs(nex, cur)
    s += t
    sub_size[cur].append(t)
  if s != N - 1 :
    sub_size[cur].append(N - 1 - s)

  return s + 1

dfs(0, -1)

fac = [1]
inv = [1]
for _ in range(N) :
  fac.append(fac[-1] * 2 % mod)
  inv.append(pow(fac[-1], mod - 2, mod))

ret = 0
for i in range(N) :
  if len(sub_size[i]) < 2 :
    continue
  
  a, b = 1, 1
  for ss in sub_size[i] :
    a *= inv[ss]
    b += fac[ss] - 1
    a %= mod
    b %= mod
  ret += (1 - a * b) % mod
  ret %= mod

ret = ret * inv[1] % mod

print(ret)