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)

from collections import deque
import sys
sys.setrecursionlimit(2*10**5)
mod=10**9+7
N=int(input())
G=[[] for i in range(N)]
for i in range(N-1):
    A,B=map(int,input().split())
    G[A-1].append(B-1)
    G[B-1].append(A-1)
Child=[[] for i in range(N)]
reached=[0 for i in range(N)]
q=deque([])
q.append(0)
while(len(q)>0):
    r=q.popleft()
    reached[r]=1
    for p in G[r]:
        if reached[p]==1:
            continue
        Child[r].append(p)
        q.append(p)
#print(Child)
#print(Parents)
Childnum=[-1 for i in range(N)]
def dp(i):
    if Childnum[i]!=-1:
        return Childnum[i]
    if len(Child[i])==0:
        Childnum[i]=0
        return 0
    res=0
    for j in Child[i]:
        res+=dp(j)+1
    Childnum[i]=res
    return res
dp(0)
#print(Childnum)
def inv(x):
    return pow(x,mod-2,mod)
inv2=[inv(pow(2,i,mod)) for i in range(N+1)]
ans=(N*(1-inv(pow(2,N-1,mod))))%mod
for r in range(N):
    tmp=0
    for p in Child[r]:
        ai=Childnum[p]+1
        tmp+=ai
        k1=(1-inv2[ai])%mod
        k2=inv2[N-1-ai]
        #print(r,p,(k1*k2)%mod)
        ans-=(k1*k2)%mod
        ans%=mod
    ai=N-1-tmp
    k1=(1-inv2[ai])%mod
    k2=inv2[N-1-ai]
    ans-=(k1*k2)%mod
    #print(r,-1,(k1*k2)%mod)
    ans%=mod
print((ans*inv(2))%mod)