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 Python (3.4.3) (judge verdict: Accepted)

from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
from pprint import pprint
from copy import deepcopy
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
from pprint import pprint


sys.setrecursionlimit(2147483647)
INF = 10 ** 15
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007

n = I()
G = [[] for _ in range(n)]
for _ in range(n - 1):
    a, b = LI()
    G[a - 1] += [b - 1]
    G[b - 1] += [a - 1]

memo = [0] * n
par = [-1] * n

def child_cnt(x):
    cnt = 1
    for y in G[x]:
        if y != par[x]:
            par[y] = x
            cnt += child_cnt(y)
    memo[x] = cnt
    return cnt

child_cnt(0)
two_mod = [1] * (n + 1)
for i in range(1, n + 1):
    two_mod[i] = two_mod[i - 1] * 2 % mod

ans = [0] * n
q = deque([0])
while q:
    u = q.pop()
    ret = two_mod[n - 1] - 1
    for v in G[u]:
        if par[u] != v:
            ret = (ret - (two_mod[memo[v]] - 1)) % mod
            q += [v]
    ret = ret - (two_mod[n - memo[u]] - 1)
    ans[u] = ret

print(sum(ans) % mod * pow(two_mod[n], mod - 2, mod) % mod)