Problem:
Score : 500 points

Problem Statement

There are N cards placed face down in a row. On each card, an integer 1 or 2 is written.

Let A_i be the integer written on the i-th card.

Your objective is to guess A_1, A_2, ..., A_N correctly.

You know the following facts:

For each i = 1, 2, ..., M, the value A_{X_i} + A_{Y_i} + Z_i is an even number.

You are a magician and can use the following magic any number of times:

Magic: Choose one card and know the integer A_i written on it. The cost of using this magic is 1.

What is the minimum cost required to determine all of A_1, A_2, ..., A_N?

It is guaranteed that there is no contradiction in given input.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq X_i < Y_i \leq N

1 \leq Z_i \leq 100

The pairs (X_i, Y_i) are distinct.

There is no contradiction in input. (That is, there exist integers A_1, A_2, ..., A_N that satisfy the conditions.)

Input

Input is given from Standard Input in the following format:

N M
X_1 Y_1 Z_1
X_2 Y_2 Z_2
\vdots
X_M Y_M Z_M

Output

Print the minimum total cost required to determine all of A_1, A_2, ..., A_N.

Sample Input 1

3 1
1 2 1

Sample Output 1

2

You can determine all of A_1, A_2, A_3 by using the magic for the first and third cards.

Sample Input 2

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5

Sample Output 2

2

Sample Input 3

100000 1
1 100000 100

Sample Output 3

99999

Solution in Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::{Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused)]
const ALPHA_SMALL: [char; 26] = [
    'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's',
    't', 'u', 'v', 'w', 'x', 'y', 'z',
];
#[allow(unused)]
const ALPHA: [char; 26] = [
    'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S',
    'T', 'U', 'V', 'W', 'X', 'Y', 'Z',
];
#[allow(unused)]
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
#[allow(unused)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(unused)]
fn read_touple<T: std::str::FromStr + Copy>() -> (T, T) {
    let v: Vec<T> = read_vec();
    assert_eq!(v.len(), 2);
    (v[0], v[1])
}
fn main() {
    input!(N: usize, M: usize);
    let mut g = vec![vec![]; N];
    for _ in 0..M {
        input!((x, y, z): (Usize1, Usize1, Usize1));
        g[x].push(y);
        g[y].push(x);
    }

    let mut visit = vec![false; N];
    let mut ans = 0;
    for i in 0..N {
        if visit[i] == false {
            bfs(N, &g, i, &mut visit);
            ans += 1;
        }
    }

    println!("{}", ans);
}

//単一視点から目的のノードまでの最短距離、ただし全ての辺の重みが等しいグラフに限る
fn bfs(n: usize, g: &Vec<Vec<usize>>, start: usize, visit: &mut Vec<bool>) {
    let mut q = std::collections::VecDeque::new();

    //init
    visit[start] = true;
    q.push_back(start);

    while let Some(now) = q.pop_front() {
        for &e in &g[now] {
            if !visit[e] {
                visit[e] = true;
                q.push_back(e);
            }
        }
    }
}