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 PyPy3 (7.3.0) (judge verdict: Accepted)

N, M = map(int, input().split())
edges = []
for _ in range(M):
    x,y,z = map(int, input().split())
    x -= 1
    y -= 1
    edges.append((x, y))

class UnionFind():
    """
    UnionFindTreeのクラス
    
    Attributes:
    n: int: ノード数
    parents: そのノードが属する木の根を格納, 根の場合は-1

    """
    def __init__(self, n):
        """ノード数nで初期化"""
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        """ノードxが属するグループの根を返す"""
        if self.parents[x] < 0:
            return x
        else:
            #経路圧縮
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        """xが属するグループとyが属するグループを併合"""
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        """xが属するグループの要素数を返す"""
        return -self.parents[self.find(x)]

    def same(self, x, y):
        """x, yが同じグループに属するかどうかを返す"""
        return self.find(x) == self.find(y)

    def members(self, x):
        """xが属するグループ内の要素をリストで返す"""
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        """全ての根に対する要素をリストで返す"""
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        """グループの総数を返す"""
        return len(self.roots())

    def all_group_members(self):
        """{root: [そのグループの要素リスト]}の辞書を返す"""
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        """printデバッグ用. root: [要素リスト]をstrで返す"""
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())


uf = UnionFind(N)
for u, v in edges:
    uf.union(u, v)
print(uf.group_count())