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)

// https://atcoder.jp/contests/abc126/tasks/abc126_c

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <tuple>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
#define INF (1e9)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
// int gcd(long a, long b) { return b ? gcd(b, a % b) : a; }
// int lcm(long a, long b) { return a * b / gcd(a, b); }

struct UnionFind {
    vector<int> data;
    UnionFind(int size) : data(size, -1) {}
    bool unionSet(int x, int y) {
        x = root(x);
        y = root(y);
        if (x != y) {
            if (data[y] < data[x])
                swap(x, y);
            data[x] += data[y];
            data[y] = x;
        }
        return x != y;
    }
    bool findSet(int x, int y) { return root(x) == root(y); }
    int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
    int size(int x) { return -data[root(x)]; }
};

// X, Yのグループが何個あるか調べればいい
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;

    UnionFind tree(n + 1);
    set<int> st;

    for (int i = 0; i < m; i++) {
        int x, y, z;
        cin >> x >> y >> z;
        tree.unionSet(x, y);
    }
    for (int i = 1; i <= n; i++) {
        st.insert(tree.root(i));
    }
    cout << st.size() << endl;

    return 0;
}