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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < n; i++)
#define forep(i, k, n) for (int i = k; i < n; i++)
const int MOD = 1000000007;

ll modpow(ll m, ll n)
{
  ll res = 1;
  while (n > 0)
  {
    if (n & 1)
    {
      res = res * m % MOD;
    }
    m = m * m % MOD;
    n >>= 1;
  }
  return res;
}

int n;
vector<vector<int>> to(200000);

ll ans = 0;
int dfs(int v, int p)
{
  vector<int> ts; // Tからvを除いて得られる各部分木のサイズ
  int res = 1;    // vを根とする部分木のサイズ
  // vの子孫である部分木のサイズを計算
  for (int nv : to.at(v))
  {
    if (nv == p)
    {
      continue;
    }
    int t = dfs(nv, v);
    res += t;
    ts.push_back(t);
  }
  // vの祖先である部分木のサイズを計算
  if (p != -1)
  {
    ts.push_back(n - res);
  }
  int now = (modpow(2, n - 1) - 1 + MOD) % MOD;
  for (int t : ts)
  {
    now = (now - modpow(2, t) + 1 + MOD) % MOD;
  }
  ans = (ans + now) % MOD;
  return res;
}

int main()
{
  cin >> n;
  rep(i, n - 1)
  {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    to.at(a).push_back(b);
    to.at(b).push_back(a);
  }
  dfs(0, -1);
  ll invx = modpow(modpow(2, n), MOD - 2);
  cout << ans * invx % MOD << endl;
}