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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
list($n, $m) = explode(" ", trim(fgets(STDIN)));
$tree = new UFT($n);
for($i = 0; $i < $m; $i++){
    list($from, $to, $_) =  explode(" ", trim(fgets(STDIN)));
    $tree->unite($from, $to);
}
for($i = 1; $i <= $n; $i++){
    $map[$tree->getRoot($i)] = true;
}
echo count($map);

class UFT{
    public $parent;
    public $size;
    public $rank;
    public function __construct($size){
        for($i = 1; $i <= $size; $i++){
            $this->size[$i] = 1;
            $this->rank[$i] = 1;
            $this->parent[$i] = 0;
        }
    }
    public function dump(){
        o("--------------");
        o($this->parent);
        o($this->size);
        o($this->rank);
    }
    public function getRoot($i){
        if($this->parent[$i] == 0){
            return $i;
        }else{
            return $this->getRoot($this->parent[$i]);
        }
    }
    public function unite($i, $j){
        $rootI = $this->getRoot($j);
        $rootJ = $this->getRoot($i);
        if($rootJ == $rootI)return false;//元から同じグループ
        if($this->rank[$rootI] > $this->rank[$rootJ])list($rootI, $rootJ) = [$rootJ, $rootI];//Rank(J)>Rank(I)に揃えておく
        $this->parent[$rootI] = $rootJ;
        if($this->rank[$rootI] == $this->rank[$rootJ]){
            $this->rank[$rootJ]++;
        }
        $this->size[$rootJ]+=$this->size[$rootI];
        $this->size[$rootI] = '*';//不要な情報となるので潰す/デバッグ出力向けに*を入れている
        $this->rank[$rootI] = '*';//同上
    }
    public function size($i){return $this->size[$this->getRoot($i)];}
    public function isUnion($i, $j){return $this->getRoot($i) == $this->getRoot($j);}
}