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 Rust (1.15.1) (judge verdict: Accepted)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut t = Tree::new(n);
    for _ in 0..n - 1 {
        let i: usize = scan(tokens);
        let j: usize = scan(tokens);
        t.add_edge(i - 1, j - 1);
    }
    println!("{}", t.solve());
}

struct Tree {
    n: usize,
    neighbors: Vec<Vec<usize>>,
    visited: Vec<bool>,
    counts: Vec<Vec<usize>>,
}

impl Tree {
    fn new(n: usize) -> Self {
        Tree {
            n: n,
            neighbors: vec![Vec::new(); n],
            visited: vec![false; n],
            counts: vec![Vec::new(); n],
        }
    }
    fn add_edge(&mut self, i: usize, j: usize) {
        self.neighbors[i].push(j);
        self.neighbors[j].push(i);
    }
    fn visit(&mut self, i: usize) -> usize {
        if self.visited[i] {
            return 0;
        }
        self.visited[i] = true;
        let mut ans = 1;
        for j in 0..self.neighbors[i].len() {
            let i1 = self.neighbors[i][j];
            let k = self.visit(i1);
            if k > 0 {
                self.counts[i].push(k);
                ans += k;
            }
        }
        self.counts[i].push(self.n - ans);
        ans
    }
    fn solve(&mut self) -> usize {
        self.visit(0);
        let power_of_two: Vec<usize> = {
            let mut power_of_two = vec![0; self.n + 1];
            power_of_two[0] = 1;
            for i in 0..self.n {
                power_of_two[i + 1] = power_of_two[i] * 2 % MODULO;
            }
            power_of_two
        };
        let mut ans = 0;
        for i in 0..self.n {
            let l = self.counts[i].len();
            if l == 1 {
                continue;
            }
            ans = (ans + power_of_two[self.n - 1]) % MODULO;
            for j in 0..l {
                let k = self.counts[i][j];
                ans = (ans + neg(power_of_two[k])) % MODULO;
            }
            ans = (ans + l - 1) % MODULO;
        }
        ans * inverse(power_of_two[self.n]) % MODULO
    }
}

const MODULO: usize = 1_000_000_007;

fn power(mut base: usize, mut exponent: usize) -> usize {
    let mut prod = 1;
    while exponent != 0 {
        if exponent & 1 == 1 {
            prod = prod * base % MODULO;
        }
        base = base * base % MODULO;
        exponent >>= 1;
    }
    prod
}

fn inverse(x: usize) -> usize {
    power(x, MODULO - 2)
}

fn neg(x: usize) -> usize {
    if x == 0 {
        0
    } else {
        MODULO - x
    }
}