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)

use proconio::{input, fastout};
use std::collections::VecDeque;

#[fastout]
fn main() {
    input!{
        (n, m): (usize, usize),
        xyz: [(usize, usize, usize); m]
    };

    let mut links: Vec<Vec<usize>> = vec![Vec::new(); n];
    for (a, b, c) in xyz {
        links[a-1].push(b-1);
        links[b-1].push(a-1);
    }

    let mut ans = 0;
    let mut used = vec![false; n];
    for i in 0..n {
        if !used[i] {
            ans += 1;
            let mut que: VecDeque<usize> = VecDeque::new();
            que.push_back(i);
            while !que.is_empty() {
                let from = match que.back() {
                    Some(x) => *x,
                    None => 0 as usize,
                };
                que.pop_back();
                if used[from] {
                    continue;
                }
                used[from] = true;
                for to in links[from].iter() {
                    if !used[*to] {
                        que.push_back(*to);
                    }
                }
            }
        }
    }
    println!("{}", ans);
}