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 <fstream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>

#define REP(i, n) for (long long i = 0; i < (n); i++) 
typedef long long ll;
static const ll MOD = 1000000007;
static const ll INF = 1000000000000000000LL;
using namespace std;

//N枚のカードが伏せられている
//Aには1か2が書かれている
//i番目に伏せられているカードをA[i]とする

// 以下の情報がM個与えられる
// A[Xi] + A[Yi] + Ziは偶数である

//Mの情報を使用しても特定できないカードは何個あるか
//(一つわかれば片方は自明、などもあるかもしれないことに注意する)

//Aの情報をまとめたい
//i 番目は i+1番目の反対、など
//UnionFindでいいのでは??

struct UnionFind {

    vector <ll> par;
    
    // new
    UnionFind(ll n):par(n, -1){}

    //初期化
    void init(ll n){par.assign(n, -1); } 


    ll root(ll x)
    {
        if(par[x] < 0)return x;
        else return par[x] = root(par[x]);//親の番号を返す(親が子になっていたら困るので取り込まれてたら代入も行う)
    }

    //同じグループに属するか
    bool issame(ll x, ll y)
    {
        //rootが同じであるか
        return root(x) == root(y);
    }

    //併合
    bool merge(ll x, ll y)
    {

        x = root(x);
        y = root(y);
        if(x == y)return false;
        if(par[x] > par[y])swap(x, y);
        //多い方に足し合わせる
        par[x] += par[y];
        //従属(ここで0以上となる)
        par[y] = x;
        return true;
    }

    ll size(ll x)
    {
        return -par[root(x)];
    }

};

int main(){
    int N,M;
    cin >> N >> M;

    UnionFind uf(N);

    int res = N;
    REP(i,M)
    {
        int x,y,z;
        cin >> x >> y >> z;
        --x;
        --y;
        //同じグループに属するか
        if(uf.issame(x,y))continue;

        //サイズ取得
        int sx = uf.size(x);
        int sy = uf.size(y);
        //xとyは一致
        uf.merge(x,y);
        --res;
        //cout << x + 1 << " " << y + 1 << endl;

    }
    cout << res << endl;

    return 0;    
}