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)

/*input

*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

using namespace __gnu_pbds;
using namespace __gnu_cxx;
using namespace std;

#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")

#define F first
#define S second
#define mp make_pair
#define append push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long int
#define pb push_back
#define print(a) cout<<a<<endl
#define gcd(a,b) __gcd((a),(b))
#define lcm(a,b) ((a)*(b))/gcd((a),(b))
#define all(x) x.begin(), x.end()
#define loop(i,n) for(int i = 0; i < n; i++)
//#define int long long int
 
const int MOD = 1e9 + 7;

int fastexpo(int x, int y) 
{ 
    int res = 1;
    x = x % MOD;
    while (y) 
    { 
        if (y & 1) res = (res*x) % MOD; 
        y = y>>1;
        x = (x*x) % MOD;   
    } 
    return res; 
}

int mod(int x, int y){
    return (x%y + y)%y;
}

void solve();

signed main(){
    
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    //cin>> t;
    while(t--){
        solve();
    }

    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}

const int MAXN = 1e5 + 10;

vector<int> graph[MAXN];
bool vis[MAXN];

void dfs(int node){
    if(vis[node])
        return;
    vis[node] = true;
    for(auto u : graph[node])
        dfs(u);
}

void solve(){
    mem(vis, 0);
    int n, m;
    cin>> n >> m;
    for(int i = 0; i < m; i++){
        int u, v, w;
        cin>> u >> v >> w;
        graph[u].append(v);
        graph[v].append(u);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++){
        if(!vis[i]){
            ans++;
            dfs(i);
        }
    }
    print(ans);
}