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;

using ll = long long;
constexpr ll mod = 1e9 + 7;

template <typename T>
void printv(const vector<T> &v) {
  int sz = v.size();
  for (int i = 0; i < sz; i++) {
    cout << v[i] << " \n"[i == sz - 1];
  }
}

ll mypow(ll a, ll b) {
  if (b == 0) return 1;
  ll tmp = mypow(a, b / 2);
  if (b % 2)
    return (((tmp * tmp) % mod) * a) % mod;
  else
    return (tmp * tmp) % mod;
}

using Graph = vector<vector<int>>;
const int MAX_N = 200010;
Graph g(MAX_N);
vector<ll> ch(MAX_N), ret(MAX_N);
vector<ll> two(MAX_N, 1);  // 2^(-i)
int n;

int dfs(int cur, int pre) {
  ch[cur] = 1;
  for (auto nxt : g[cur]) {
    if (nxt == pre) continue;
    ch[cur] += dfs(nxt, cur);
  }
  if (g[cur].size() <= 1) {
    return ch[cur];
  }
  ret[cur] = two[n - 1] - 1;
  for (auto nxt : g[cur]) {
    if (nxt == pre) continue;
    ret[cur] = (ret[cur] + mod - two[ch[nxt]] + 1) % mod;
  }
  ll r = n - ch[cur];
  ret[cur] = (ret[cur] + mod - two[r] + 1) % mod;
  return ch[cur];
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  for (ll i = 1; i < MAX_N; i++) {
    two[i] = two[i - 1] * 2 % mod;
  }
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  ll sum = 0;
  for (int i = 0; i < n; i++) sum = (sum + ret[i]) % mod;
  ll tmp = mypow(2LL, mod - 2);
  tmp = mypow(tmp, ll(n));
  sum = sum * tmp % mod;
  cout << sum << endl;
}