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 Python (3.4.3) (judge verdict: Accepted)

class UnionFind:
    def __init__(self, size: int):
        self.__size = size
        self.__root = [-1] * size

    def find(self, x: int)->int:
        if self.__root[x] < 0:
            return x
        self.__root[x] = self.find(self.__root[x])
        return self.__root[x]

    def union(self, x: int, y: int)->int:
        rx, ry = self.find(x), self.find(y)

        if rx == ry:
            return

        if self.__root[ry] < self.__root[rx]:
            rx, ry = ry, rx

        self.__root[rx] += self.__root[ry]
        self.__root[ry] = rx

    def same(self, x: int, y: int)->bool:
        return self.find(x) == self.find(y)


def one_or_two(N: int, M: int, queries: list)->int:
    uf = UnionFind(N)

    for x, y, _ in queries:
        uf.union(x-1, y-1)

    return len(set(uf.find(i) for i in range(N)))


if __name__ == "__main__":
    M = 0
    N, M = map(int, input().split())
    queries = [tuple(int(s) for s in input().split()) for _ in range(M)]

    ans = one_or_two(N, M, queries)
    print(ans)