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)

#pragma GCC optimize("Ofast")
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"

using namespace std;

using u8 = uint8_t;
using u16 = uint16_t;
using u32 = uint32_t;
using u64 = uint64_t;
using i8 = int8_t;
using i16 = int16_t;
using i32 = int32_t;
using i64 = int64_t;

constexpr char newl = '\n';
constexpr double eps = 1e-10;

#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define F0R(i,b) FOR(i,0,b)
#define RFO(i,a,b) for (int i = ((b)-1); i >=(a); i--)
#define RF0(i,b) RFO(i,0,b)
#define fi first
#define se second
#define debug(x) cout << #x << ": " << x << '\n';
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()

template<typename T1, typename T2> inline void chmin(T1& a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1& a, T2 b) { if (a < b) a = b; }
template<class T> void Print(vector<T> v) {
  F0R(i, v.size()) {
    cout << v[i] << ' ';
  }
  cout << newl;
}

#if 1

constexpr int MOD = 1e9 + 7; // 1e9 + 7

template<int MOD>
struct Modint {
  long long val;
  constexpr Modint(long long v = 0) noexcept : val(v% MOD) {
    if (val < 0) val += MOD;
  }
  constexpr int getmod() { return MOD; }
  constexpr Modint operator - () const noexcept {
    return val ? MOD - val : 0;
  }
  constexpr Modint operator + (const Modint& r) const noexcept { return Modint(*this) += r; }
  constexpr Modint operator - (const Modint& r) const noexcept { return Modint(*this) -= r; }
  constexpr Modint operator * (const Modint& r) const noexcept { return Modint(*this) *= r; }
  constexpr Modint& operator += (const Modint& r) noexcept {
    val += r.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  constexpr Modint& operator -= (const Modint& r) noexcept {
    val -= r.val;
    if (val < 0) val += MOD;
    return *this;
  }
  constexpr Modint& operator *= (const Modint& r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr bool operator == (const Modint& r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator != (const Modint& r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream& operator << (ostream& os, const Modint<MOD>& x) noexcept {
    return os << x.val;
  }
  friend constexpr istream& operator >> (istream& is, Modint<MOD>& x) noexcept {
    return is >> x.val;
  }
  
  constexpr Modint<MOD> pow(long long n) const noexcept {
    if (!n) return 1;
    auto t = pow(n >> 1);
    t *= t;
    if (n & 1) t *= *this;
    return t;
  }
  friend constexpr Modint<MOD> modpow(const Modint<MOD>& a, long long n) noexcept {
    return a.pow(n);
  }

  friend constexpr Modint<MOD> Ncr(Modint<MOD> a, int n) {
    Modint<MOD> p = 1, q = 1;

    while (n > 0) {
      p *= a;
      a -= 1;
      q *= n--;
    }

    return p / q;
  }

  // for prime mod
  constexpr Modint& operator /= (const Modint& r) noexcept {
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b; swap(a, b);
      u -= t * v; swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0) val += MOD;
    return *this;
  }
  constexpr Modint operator / (const Modint& r) const noexcept { return Modint(*this) /= r; }
};
template<int MOD>
using mint = Modint<MOD>;


// INSERT ABOVE HERE
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int N;
  cin >> N;
  vector<vector<int>> G(N);
  F0R(i, N - 1) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }

  // 頂点ごとに、白として含まれるのは何通りであるか数える

  // ・頂点は白
  // ・内部になる（黒にはさまれる）
  //   = 全 - 内部にならない（余事象）

  // 内部になる = 頂点からのびている部分木のうち2つ以上に黒が含まれる
  // 内部にならない = 高々1方向にしか黒が含まれない

  // 全 - 内部にならない
  // 全：2^(N-1) 頂点は白なので、自由にできる頂点はN-1
  // 内部にならない：
  // ・全部白
  //   1
  // ・ちょうど1個の部分木のみに黒がある
  //   個Ciの部分木のサイズをTCiと置く
  //   
  //   Σ(2^(TCi)-1)
  //   i
  //   2^(TCi) ... 部分木それぞれの頂点が白か黒か
  //   -1      ... 黒は一つ必要なので、全部白のパターンを引く

  vector<mint<MOD>> two(N+1);
  two[0] = 1;
  F0R(i, N) {
    two[i + 1] = two[i] * 2;
  }

  vector<int> vs(N);

  mint<MOD> result = (two[N - 1] - 1) * N;

  auto dfs = [&](auto dfs, int v, int p)->int {
    int r = 1;
    for (auto c : G[v]) {
      if (c == p) continue;

      int cv = dfs(dfs, c, v);
      result -= two[cv] - 1;
      r += cv;
    }

    // 親側の部分木
    int pv = N - r;
    result -= two[pv] - 1;
    return r;
  };
  dfs(dfs, 0, -1);

  result /= two[N];
  cout << result;
}
#endif