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>
#define REP(i, n) for(int i=0; i<(int)(n); ++i)
#define FOR(i, s, n) for(int i=(int)(s); i<(int)(n); ++i)
#define ALL(n) (n).begin(), (n).end()
#define pb push_back

using namespace std;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> mat;

const ll MAX_N = 100000;
const ll INF = 1e9;
const ll MOD = 998244353;

class UnionFindTree{
private:
    vector<int> parent;
    vector<int> rank;
public:
    int n;
    UnionFindTree(int n);
    int root(int x);
    void unite(int x, int y);
    bool isSame(int x, int y);
};

UnionFindTree::UnionFindTree(int n_): parent(n_), rank(n_, 0){
    n = n_;
    // 初期化:すべてのノードの親を自分にする
    REP(i, n) parent[i] = i;
}

int UnionFindTree::root(int x){
    // 経路圧縮を行う
    if (parent[x] == x) return x;
    else return parent[x] = root(parent[x]);
}

void UnionFindTree::unite(int x, int y){
    // 木の結合:高さが低い木を高い木の下にぶら下げる
    int rx = root(x);
    int ry = root(y);
    if (rx == ry) return;
    else {
        if (rank[rx] < rank[ry]){
            parent[rx] = ry;
        }else{
            parent[ry] = rx;
            if (rank[rx] == rank[ry]) rank[rx]++;
        }
    }
}

bool UnionFindTree::isSame(int x, int y){
    return root(x) == root(y);
}

int main(){
    int N, M; cin >> N >> M;
    set<int> roots;
    UnionFindTree uf(N);
    REP(i, M){
        int x, y, z; cin >> x >> y >> z;
        uf.unite(x-1, y-1);
    }
    REP(i, N){
        roots.insert(uf.root(i));
    }
    cout << roots.size() << endl;
}