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)

from sys import stdin
f_i = stdin

N = int(f_i.readline())

adj = [[] for i in range(N)]
for i in range(N - 1):
    A, B = map(int, f_i.readline().split())
    A -= 1
    B -= 1
    adj[A].append(B)
    adj[B].append(A)

mod = 10 ** 9 + 7
pow2 = [(2 ** i) % mod for i in range(N + 1)]

from sys import setrecursionlimit
setrecursionlimit(10 ** 6)

ans = 0
def dfs(node, parent):
    global ans
    if len(adj[node]) == 1:
        if adj[node][0] == parent:
            ans += (pow2[N-1] - 1)
            ans %= mod
            return 1
    cnt = 1
    for child in adj[node]:
        if child == parent:
            continue
        cnt += dfs(child, node)
    if node != 0:
        ans += (pow2[cnt] - 1) * (pow2[N-cnt] - 1)
        ans %= mod
    return cnt

dfs(0, None)

ans = (ans - N * pow2[N-1] + pow2[N] - 1) % mod
ans = ans * pow(pow2[N], mod - 2, mod) % mod

print(ans)