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;

#define int long long
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rep1(i,n) for(int i=1;i<=(int)(n);++i)
#define rep11(i,n) for(int i=1;i<(int)(n);++i)
#define repo(i,o,n) for(int i=o;i<(int)(n);++i)
#define repm(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define sperase(v,n) (v).erase(remove(all(v), n), (v).end())
#define vdelete(v) (v).erase(unique(all(v)), (v).end())
#define pb(n) push_back(n)
#define mp make_pair
#define MOD 1000000007
#define INF 9223372036854775807

struct UnionFind {
#define MAX 100001
    int ElementCount[MAX],Adress[MAX];
    void Init() {rep(i,MAX) ElementCount[i] = 1,Adress[i] = i;}
    void unite(int a ,int b) {
        int a_root = find(a),b_root = find(b);
        if (a_root != b_root) {
            ElementCount[a_root] += ElementCount[b_root];
            Adress[b_root] = a_root;
        }
    }
    int find(int a) {
        if (a == Adress[a]) return Adress[a];
        else {
            Adress[a] = find(Adress[a]);
            return Adress[a];
        }
    }
    int size(int a) {
        int root = find(a);
        return ElementCount[root];
    }
    /*
     * ElementCount : 属するグループの要素数
     * Adress : rootの番号
     * unite : 併合
     * find : aのrootを返す
     * size : aのrootのElementCountを返す
     * */
};


int n,m,x,y,z;
UnionFind UF;
set<int> s;

signed main() {
    UF.Init();
    cin >> n >> m;
    rep(i,m) {
        cin >> x >> y >> z;
        UF.unite(x,y);
    }
    rep1(i,n) s.insert(UF.find(i));
    cout << s.size() << endl;
}