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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <numeric>
#include <ctime>
#include <complex>
#include <bitset>
#include <random>
#include <climits>
#include <stack>

using namespace std;

typedef long long ll;
typedef long double ld;

#define int ll
#define double ld
#define loop(i, n) for(int i = 0; i < (int)n; ++i)
#define loop1(i, n) for(int i = 1; i <= (int)n; ++i)
#define F first
#define S second
#define pb push_back
#define pi pair <int, int>
#define all(x) begin(x), end(x)
#define ti tuple <int, int, int>
#define Point Vect
#define no {cout << 0; return;}
#define mkp make_pair
#define mkt make_tuple
#define cerr if(0) cerr

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

int binpow(int x, int deg) {
    int val = 1;
    while (deg) {
        if (deg & 1)
            val = val * x % MOD;
        x = x * x % MOD;
        deg >>= 1;
    }
    return val;
}

vector <int> g[N];
int ans, rev2, sz[N];

void dfs(int v, int p, int n) {
    sz[v] = 1;
    int mns = 0;
    for (int u : g[v])
        if (u != p) {
            dfs(u, v, n);
            sz[v] += sz[u];
            mns = (mns + binpow(2, sz[u]) - 1) % MOD;
        }
    mns = (mns + binpow(2, n - sz[v])) * 2 % MOD;
    ans = (ans + (binpow(2, n) - mns + MOD) % MOD * binpow (rev2, n + 1)) % MOD;
}

void solve() {
    rev2 = binpow(2, MOD - 2);
    int n; cin >> n;
    loop(i, n - 1) {
        int u, v;
        cin >> u >> v;
        g[v].pb(u);
        g[u].pb(v);
    }
    dfs(1, 1, n);
    cout << ans;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
}