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 (1.15.1) (judge verdict: Accepted)

use std::io::{self, BufRead};
use std::collections::BTreeSet;

fn bfs(es: &Vec<Vec<usize>>, rest: &mut BTreeSet<usize>, n: usize) {
  rest.remove(&n);
  for &c in &es[n] {
    if rest.contains(&c) {
      bfs(es, rest, c);
    }
  }
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let (n, m) = (xs[0], xs[1]);
  let mut es: Vec<Vec<usize>> = vec![Vec::new(); n];
  for _ in 0..m {
    let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
    let (x, y) = (xs[0] - 1, xs[1] - 1);
    es[x].push(y);
    es[y].push(x);
  }
  let mut rest: BTreeSet<usize> = (0..n).collect();
  let mut count: usize = 0;
  while !rest.is_empty() {
    let n = rest.iter().next().unwrap().clone();
    bfs(&es, &mut rest, n);
    count += 1;
  }
  println!("{}", count);
}