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: Time Limit Exceeded)

import sys
input=lambda: sys.stdin.readline().rstrip()
sys.setrecursionlimit(200000)
mod=10**9+7
n=int(input())
edge=[[] for i in range(n)]
for i in range(n-1):
  a,b=map(int,input().split())
  edge[a-1].append(b-1)
  edge[b-1].append(a-1)
inf=float("inf")
Par=[inf]*n
Par[0]=-1
Chk=[0]
while Chk:
  c=Chk.pop()
  for next in edge[c]:
    if Par[next]==inf:
      Par[next]=c
      Chk.append(next)

C=[-1]*n
def ch(x):
  ret=0
  if x!=0 and len(edge[x])==1:
    C[x]=0
    return C[x]
  else:
    for e in edge[x]:
      if e==Par[x]:
        continue
      else:
        ret+=ch(e)+1
    C[x]=ret
    return C[x]
ch(0)

H=[0]*n
H[0]=1
H[1]=pow(2,mod-2,mod)
for i in range(2,n):
  H[i]=(H[i-1]*H[1])%mod

ans=0
for i in range(n):
  if len(edge[i])==1:
    continue
  else:
    A=[]
    for e in edge[i]:
      if e==Par[i]:
        A.append(n-1-C[i])
      else:
        A.append(C[e]+1)
    cur=1+(len(edge[i])-1)*H[-1]
    for a in A:
      cur=(cur-H[n-1-a])%mod
    ans=(ans+cur)%mod
print((ans*H[1])%mod)