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 <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <set>
#include <utility>
#include <cstdlib>
#include <queue>
#include <stack>
#include <iomanip>
#include <cstdio>
#include <map>

using namespace std;

using ll = long long;
using P = pair<ll,ll>;

#define rep(i,x) for(ll i = 0; i < (ll)(x); i++)
#define rrep(i,x) for(ll i = ((ll)(x) - 1); i >= 0; i--)

#define _upgrade ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define all(x) (x).begin(), (x).end()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

template<class T1, class T2> void chmin(T1 &a, T2 b){if(a>b)a=b;}
template<class T1, class T2> void chmax(T1 &a, T2 b){if(a<b)a=b;}

//ll gcd(ll a, ll b){return b?gcd(b,a%b):a;}
//ll lcm(ll x, ll y) {return x / gcd(x, y) * y;}

int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};

const ll mod = 1e9 + 7;
const ll INF = 1LL << 60;
const double PI = 3.14159265358979323846;


// ****************************************CODE***************************************//











struct UnionFind {

        vector<int> par;

        UnionFind(int n) : par(n) {
                for(int i = 0; i < n; i++) par[i] = i;
        }

        int root(int x){
                if(par[x] == x) return x;
                return par[x] = root(par[x]);
        }

        void Union(int x, int y){
                int nx = root(x);
                int ny = root(y);

                if(nx == ny) return;

                par[nx] = ny;
        }
};

int main(){
        int n,m;
        cin >> n >> m;

        UnionFind tree(n);
        for(int i = 0; i < m; i++)
        {
                int X, Y, Z;
                cin >> X >> Y >> Z;
                X--;
                Y--;

                tree.Union(X, Y);
        }

        int sum = 0;
        for(int i = 0; i < n; i++){
                if(i == tree.root(i)) sum++;
        }

        cout << sum << endl;
}