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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

vector <ll> id;

ll find(ll i){
	if (id[i] < 0) return i;
	return id[i] = find(id[i]);
}

ll unio(ll i,ll j){
	ll a = find(i);
	ll b = find(j);
	if (a == b) return 0;
	if (id[b] < id[a]) swap(a,b);
	id[a] += id[b];
	id[b] = a;
	return id[a];
}

int main(){
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL);
    ll n,m;
    cin >> n >> m;
    id.resize(n+1,-1);
    for (int z=0;z<m;z++){
    	ll a,b,c;
    	cin >> a >> b >> c;
    	unio(a,b);
    }
    ll ans = 0;
    for (int z=1;z<=n;z++){
    	if (id[z] < 0) ans++;
    }
    cout << ans << endl;
    cin >> n;
}