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 (2.4.0) (judge verdict: Accepted)

import collections, math, bisect

local = False
if local:
    file = open("input.txt", "r")
    import time

def inp():
    if local:
        return file.readline().rstrip()
    else:
        return input().rstrip()

def ints():
    return [int(_) for _ in inp().split()]


if local:
    start=time.time()

N,M = ints()

connection = [dict() for i in range(N)]

for i in range(1, M+1):
    x,y,z = ints()

    connection[x-1][y-1] = True
    connection[y-1][x-1] = True

ans = 0
visited = set()

for i in range(N):
    if i in visited:
        continue

    ans+=1
    stck = [i]
    visited.add(i)

    while len(stck)>0:
        vertex = stck.pop(-1)

        for nextVertex in connection[vertex]:
            if nextVertex not in visited:
                visited.add(nextVertex)
                stck.append(nextVertex)

print(ans)


if local:
    fin = (time.time()-start)*1000
    print("{:.2f}".format(fin) + "ms")