Problem:
Score : 900 points

Problem Statement

We have a tree with N vertices.
The vertices are numbered 0 through N - 1, and the i-th edge (0 ≤ i < N - 1) comnnects Vertex a_i and b_i.
For each pair of vertices u and v (0 ≤ u, v < N), we define the distance d(u, v) as the number of edges in the path u-v.

It is expected that one of the vertices will be invaded by aliens from outer space.
Snuke wants to immediately identify that vertex when the invasion happens.
To do so, he has decided to install an antenna on some vertices.

First, he decides the number of antennas, K (1 ≤ K ≤ N).
Then, he chooses K different vertices, x_0, x_1, ..., x_{K - 1}, on which he installs Antenna 0, 1, ..., K - 1, respectively.
If Vertex v is invaded by aliens, Antenna k (0 ≤ k < K) will output the distance d(x_k, v).
Based on these K outputs, Snuke will identify the vertex that is invaded.
Thus, in order to identify the invaded vertex no matter which one is invaded, the following condition must hold:

For each vertex u (0 ≤ u < N), consider the vector (d(x_0, u), ..., d(x_{K - 1}, u)). These N vectors are distinct.

Find the minumum value of K, the number of antennas, when the condition is satisfied.

Constraints

2 ≤ N ≤ 10^5

0 ≤ a_i, b_i < N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_0 b_0
a_1 b_1
:
a_{N - 2} b_{N - 2}

Output

Print the minumum value of K, the number of antennas, when the condition is satisfied.

Sample Input 1

5
0 1
0 2
0 3
3 4

Sample Output 1

2

For example, install an antenna on Vertex 1 and 3.
Then, the following five vectors are distinct:

(d(1, 0), d(3, 0)) = (1, 1)

(d(1, 1), d(3, 1)) = (0, 2)

(d(1, 2), d(3, 2)) = (2, 2)

(d(1, 3), d(3, 3)) = (2, 0)

(d(1, 4), d(3, 4)) = (3, 1)

Sample Input 2

2
0 1

Sample Output 2

1

For example, install an antenna on Vertex 0.

Sample Input 3

10
2 8
6 0
4 1
7 6
2 3
8 6
6 9
2 4
5 8

Sample Output 3

3

For example, install an antenna on Vertex 0, 4, 9.

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
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)*}
    };
}

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)*}
    };
}

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>>()
    };

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

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

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

const INF: i32 = 1 << 25;

fn dfs(g: &[Vec<usize>], v: usize, par: usize) -> [i32; 2] {
    let mut ch = vec![];
    for &w in &g[v] {
        if par == w { continue; }
        ch.push(w);
    }
    if ch.len() == 0 {
        return [0, 1];
    }
    let mut ret = [1; 2];
    let mut ma = 0;
    for &w in &ch {
        let sub = dfs(g, w, v);
        for j in 0..2 {
            ret[j] += sub[j];
            if sub[j] == -INF {
                ret[j] = -INF;
            }
        }
        ma = max(ma, sub[1] - sub[0]);
    }
    if ch.len() >= 2 {
        ret[1] = -INF;
        ret[0] += ma;
    } else {
    }
    //eprintln!("dfs({}) = {:?}", v, ret);
    ret
}

// 0: parent doesn't need to have antennas
// 1: parent needs to have antennas
fn dfs2(g: &[Vec<usize>], v: usize, par: usize) -> [i32; 2] {
    let mut ch = vec![];
    for &w in &g[v] {
        if par == w { continue; }
        ch.push(w);
    }
    if ch.len() == 0 {
        return [0, 0];
    }
    if ch.len() == 1 {
        let mut sub = dfs2(g, ch[0], v);
        if sub[0] == 0 {
            sub[0] = 1;
        }
        return sub;
    }
    let mut ans = [0; 2];
    let mut vecs = vec![vec![]; 2];
    for &w in &ch {
        let sub = dfs2(g, w, v);
        for i in 0..2 {
            vecs[i].push(sub[1]);
        }
    }
    for i in 0..2 {
        vecs[i].sort();
        for j in i..vecs[i].len() {
            vecs[i][j] = max(vecs[i][j], 1);
        }
        for &a in &vecs[i] {
            ans[i] += a;
        }
    }
    //eprintln!("{} -> {:?}", v, ans);
    ans
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }
    input! {
        n: usize,
        ab: [(usize, usize); n - 1],
    }
    let mut g = vec![vec![]; n];
    for &(a, b) in &ab {
        g[a].push(b);
        g[b].push(a);
    }
    let ans;
    if g.len() == 2 {
        ans = 1;
    } else {
        let mut r = 0;
        for i in 0..n {
            if g[i].len() >= 2 {
                r = i;
                break;
            }
        }
        ans = dfs2(&g, r, n)[1];
    }
    puts!("{}\n", ans);
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}