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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define M int (1e9+7)
struct card {
	int x, y;
};
bool cmp(card a, card b) {
	if (a.x != b.x)
		return a.x < b.x;
	else
		return a.y > b.y;
}
int main() {
	int n, m;
	cin >> n >> m;
	vector<card>a(m);
	for (int i = 0; i < m; i++) {
		int e;
		cin >> a[i].x >> a[i].y >> e;
		a[i].x--;
		a[i].y--;
	}
	sort(a.begin(), a.end(), cmp);
	vector<int>d(n, -1);
	for (int i = m-1; i >= 0; i--) {
		if (d[a[i].x] == -1 && d[a[i].y] == -1) {
			d[a[i].x] = 1;
			d[a[i].y] = 0;
		}
		else if (d[a[i].x] != -1 && d[a[i].y] == -1)
			d[a[i].y] = 0;
		else if (d[a[i].x] == -1 && d[a[i].y] != -1)
			d[a[i].x] = 0;
		//else if (d[a[i].x] != -1 && d[a[i].y] != -1)
			//d[a[i].x] = d[a[i].y] = 0;
	}
	int ans = 0;
	for (int i = 0; i < n; i++)
		ans += (d[i] != 0) ? 1 : 0;
	cout << ans << endl;
	return 0;
}