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>
#include <boost/range/algorithm.hpp>
#include <boost/range/numeric.hpp>
#include <boost/range/irange.hpp>
#include <boost/range/adaptor/indexed.hpp>
 
using namespace std;
using namespace boost::adaptors;
using namespace std::string_literals;
using ll = int64_t;
using vecint = vector<int>;
using vecll = vector<ll>;
using boost::irange;
constexpr ll MOD = 1000000007;

// a^-1 mod p
ll inv(ll a,ll p){
  return ( a == 1 ? 1 : (1 - p*inv(p%a,a)) / a + p );
}

ll pow2(ll n) {
  if (n == 0) return 1;
  else if ((n%2) == 1) return (2 * pow2(n-1)) % MOD;
  else {
    ll h = pow2(n/2);
    return h * h % MOD;
  }
}

ll calc(ll n) {
  ll p2 = pow2(n);
  ll pi2 = inv(p2, MOD);
  return (p2 + MOD - 1) * pi2 % MOD;
}

ll solve(const vector<vecint>& t, vecll& ex, int i, int p) {
  vecll cp;
  ll cntc = 0;
  for(int j:t[i]) {
    if(j==p) continue;
    ll cs = solve(t,ex,j,i);
    cp.push_back(calc(cs));
    cntc += cs;
  }
  cp.push_back(calc(t.size() - cntc - 1));
  vecll fwd;
  vecll bwd;
  ll imul = 1;
  for(auto&& e:cp) {
    fwd.push_back(imul);
    imul *= (1 + (MOD - e)) % MOD;
    imul %= MOD;
  }
  boost::reverse(cp);
  imul = 1;
  for(auto&& e:cp) {
    bwd.push_back(imul);
    imul *= (1 + (MOD - e)) % MOD;
    imul %= MOD;
  }
  boost::reverse(cp);
  boost::reverse(bwd);
  ll ans = 1 + (MOD - imul);
  ans %= MOD;
  for(auto&& j:irange(0ul,cp.size())) {
    ll tmp = fwd[j] * bwd[j];
    tmp %= MOD;
    tmp *= cp[j];
    tmp %= MOD;
    ans += (MOD - tmp) % MOD;
    ans %= MOD;
  }
  ans *= inv(2, MOD);
  ans %= MOD;
  ex[i] = ans;
  return cntc + 1;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;
  cin>>n;
  vector<vecint> t(n);
  for(auto&& i:irange(0l,n-1)) {
    int a,b;
    cin>>a>>b;
    --a;--b;
    t[a].push_back(b);
    t[b].push_back(a);
  }
  vecll ex(n, 0);
  solve(t, ex, 0, -1);
  ll ans = 0;
  for(auto&& e : ex) {
    ans += e;
    ans %= MOD;
  }
  cout<<ans<<endl;
  return 0;
}