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: Runtime Error)

#https://atcoder.jp/contests/abc149/submissions/9700744
#
#[mod]
#https://atcoder.jp/contests/abc149/submissions/9228192
#https://qiita.com/drken/items/3b4fdf0a78e7a138cd9a

#mod
MOD = 7 + 10 ** 9
def power(x, y)
  if y == 0
    1
  elsif y == 1
    x % MOD
  elsif y % 2 == 0
    power(x, y/2)**2 % MOD
  else
    power(x, (y-1)/2)**2 * x % MOD
  end
end
def mul(a, b)
  ((a % MOD) * (b % MOD)) % MOD
end
def div(a, b)
  mul(a, power(b, MOD-2))
end


def dfs(c, p=-1)
  res = 1
  ts = []
  
  @data[c].each do |nc|
    next if nc == p
    t = dfs(nc, c)
    res += t
    ts << t
  end
  
  if p != -1
    ts << N - res
  end
  
  now = power(2, (N - 1)) - 1
  ts.each do |t|
    now -= power(2, t) - 1
  end
  @ans += now
  return res
end


N = gets.to_i
AB = (N-1).times.map{gets.split.map(&:to_i)}
@data = Array.new(N){Array.new()}

AB.each do |a, b|
  @data[a - 1] << b - 1
  @data[b - 1] << a - 1
end

@ans = 0
dfs(0)

p div(@ans, power(2, N))