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;

const int MAXN = 2e5+5;
const int INF = 1e9+7;
#define ll long long
#define pb push_back
#define endl '\n'
#define fast_io ios_base::sync_with_stdio(false); cin.tie(NULL);
#define F first
#define S second

ll poww(ll a, ll b, ll md) {
    return (!b ? 1 : (b & 1 ? a * poww(a * a % md, b / 2, md) % md : poww(a * a % md, b / 2, md) % md));
}

ll MOD(ll a){
    return ((a%INF)+INF)%INF;
}

ll inv(ll a){
    return poww(a,INF-2,INF);
}

int x[MAXN],y[MAXN];
vector<int> adjList[MAXN];
bool visited[MAXN];

void dfs(int start){
    visited[start] = true;
    for(auto u : adjList[start]) if(!visited[u]) dfs(u);
}

int main()
{
    fast_io;
    //cout << fixed << setprecision(15);

    int n,m;cin >> n >> m;
    for(int i=0;i<m;++i){
        int z;
        cin >> x[i] >> y[i] >> z;
        adjList[x[i]].pb(y[i]);
        adjList[y[i]].pb(x[i]);
    }
    int comp=0;
    for(int i=1;i<=n;++i) if(!visited[i]) dfs(i),++comp;
    cout << comp;
}