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: Wrong Answer)

n, m = map(int, input().split())
xyz = [list(map(int, input().split())) for _ in range(m)]
type_a = []
type_b = []
for i in range(m):
    if xyz[i][2] % 2 == 1:
        # z が奇数 = ax と ay は奇遇が異なる
        if xyz[i][0] in type_a:
            type_b.append(xyz[i][1])
        elif xyz[i][0] in type_b:
            type_a.append(xyz[i][1])
        else:
            type_b.append(xyz[i][1])
    else:
        # z が偶数 = ax と ay は奇遇が同じ
        if xyz[i][0] in type_a:
            type_a.append(xyz[i][1])
        elif xyz[i][0] in type_b:
            type_b.append(xyz[i][1])
        else:
            type_a.append(xyz[i][1])
uniq_a = list(set(type_a))
uniq_b = list(set(type_b))
print(n - len(uniq_a) - len(uniq_b))