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 Ruby (2.3.3) (judge verdict: Accepted)

N = gets.to_i
ABs = readlines.map{|l| l.split.map(&:to_i)}
M = 10**9 + 7
I2 = (M + 1)/2

@g = ABs.each_with_object(Array.new(N + 1){[]}){|(a, b), h| h[a] << b; h[b] << a}
st, pr, od = [1], [nil, 0], []
while !st.empty?
  od.unshift(u = st.pop)
  @g[u].each do |v|
    next if v == pr[u]
    pr[v] = u
    st << v
  end
end
sz = od.each_with_object(Array.new(N + 1, 1)){|v, o| o[pr[v]] += o[v]}
i2 = N.times.each_with_object([1]){|i, o| o << o.last * I2 % M}
I2N = i2[N]
puts (sz.drop(1).inject(0){|s, x| (s + (1 - i2[x] - i2[N - x] + I2N)) % M} - N*I2 - I2N + 1) % M