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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<cstdint>


int64_t const MOD = 1e9+7;
int const MAXN = 2e5;
int N;

template<int64_t Prime>
struct mod_int {
  int64_t v_;
  mod_int(int64_t x=0) : v_(x) { normalize(); }
  void normalize() { ((v_ %= Prime) += Prime) %= Prime; }
  mod_int operator+=(mod_int const& r){ (v_ += r.v_) %= Prime; return *this; }
  mod_int operator-=(mod_int const& r){ (v_ += Prime - r.v_) %= Prime; return *this; }
  mod_int operator*=(mod_int const& r){ (v_ *= r.v_) %= Prime; return *this; }
  mod_int operator+(mod_int const& r) { mod_int res(*this); return res += r; }
  mod_int operator-(mod_int const& r) { mod_int res(*this); return res -= r; }
  mod_int operator*(mod_int const& r) { mod_int res(*this); return res *= r; }
  mod_int pow(int x) const {
    int64_t res = 1, v = v_;
    while(x > 0) {
      if(x&1) (res *= v) %= Prime;
      x/=2; (v *= v) %= Prime;
    }
    return mod_int(res);
  }
  mod_int inv() const { return pow(Prime-2); }
  //operator int64_t() const { return v_; }
};
typedef mod_int<MOD> mint;

std::vector<int> G[MAXN];

int nc[MAXN];
mint dfs1(int v, int p) {
  int num = 1;
  mint res;
  for(auto c: G[v]) if(c != p) {
    res += dfs1(c, v);
    num += nc[c];
  }
  res += (mint(2).pow(num)-1)*(mint(2).pow(N - num)-1);
  nc[v] = num;
  return res;
}

int main() {
  std::cin >> N;
  for(int i = 0; i < N-1; ++i) {
    int A, B; std::cin >> A >> B; --A; --B;
    G[A].push_back(B);
    G[B].push_back(A);
  }
  mint ans = (dfs1(0, -1) - 1)*mint(2).pow(N).inv();
  ans -= mint(N)*mint(2).inv() - 1;
  std::cout << ans.v_ << std::endl;
  return 0;
}