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)

#[allow(unused_macros)]
macro_rules! input_byline {
    ($($r:tt)* ) => {
        let stdin = std::io::stdin();
        let mut q = std::collections::VecDeque::new();
        let mut next = || {
            if q.is_empty() {
                let mut s = String::new();
                stdin.read_line(&mut s);
                for v in s.split_whitespace() {
                    q.push_back(v.to_string());
                }
            }
            q.pop_front().unwrap()
        };
        input_inner!{next, $($r)*}
    };

}

#[allow(unused_macros)]
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

trait Printable {
    fn p(&self);
}

impl<T: std::fmt::Display> Printable for T {
    fn p(&self) {
        println!("{}", self);
    }
}

impl<T: std::fmt::Display> Printable for [T] {
    fn p(&self) {
        for i in 0..self.len() {
            if i != 0 {
                print!(" ");
            }
            print!("{}", self[i]);
        }
        println!("");
    }
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BinaryHeap, HashMap};

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;
#[allow(dead_code)]
const M: usize = 1e9 as usize + 7;

fn dfs(
    cur: usize,
    par: Option<usize>,
    graph: &Graph,
    nodes: &mut Vec<usize>,
    pars: &mut Vec<Option<usize>>,
) -> usize {
    let mut res = 0;
    pars[cur] = par;
    for &next in &graph[cur] {
        if par.is_some() && par.unwrap() == next {
            continue;
        }
        res += dfs(next, Some(cur), graph, nodes, pars);
    }
    nodes[cur] = res + 1;
    nodes[cur]
}

fn gcd(a: usize, b: usize) -> usize {
    if b != 0 {
        gcd(b, a % b)
    } else {
        a
    }
}

fn lcm(a: usize, b: usize) -> usize {
    a / gcd(a, b) * b
}

pub fn extgcd(a: i64, b: i64) -> (i64, i64, i64) {
    if b == 0 {
        (a, 1, 0)
    } else {
        let (gcd, x, y) = extgcd(b, a % b);
        (gcd, y, x - (a / b) * y)
    }
}

pub fn mod_inverse(a: usize, m: usize) -> usize {
    let (_, x, _) = extgcd(a as i64, m as i64);
    (x + m as i64) as usize % m
}

fn main() {
    input! {
        n: usize,
        edges: [(usize1, usize1); n-1],
    }

    let mut graph = vec![vec![]; n];

    for &(a, b) in &edges {
        graph[a].push(b);
        graph[b].push(a);
    }

    let mut nodes = vec![0; n];
    let mut pars = vec![None; n];

    dfs(0, None, &graph, &mut nodes, &mut pars);

    let mut n2 = vec![0; n + 1];
    n2[0] = 1;
    for i in 1..n + 1 {
        n2[i] = (n2[i - 1] * 2) % M;
    }

    let mut total = 0;
    for i in 0..n {
        let mut sum = (n2[n - 1] + M - 1) % M;
        for &next in &graph[i] {
            let childs = if pars[i].is_some() && pars[i].unwrap() == next {
                nodes[0] - nodes[i]
            } else {
                nodes[next]
            };
            sum = (M + sum - n2[childs] + 1) % M;
        }
        total = (total + sum) % M;
    }
    println!("{}", (total * mod_inverse(n2[n], M)) % M);
}