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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef vector<LL> VI;
typedef vector<VI> VVI;
typedef pair<LL, LL> PII;
typedef vector<string> VS;
inline LL toInt(string s) {LL v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
#define PB push_back
#define MP make_pair
#define SZ(a) LL((a).size())
#define EACH(x,c) for(auto x: (c))
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define ALL(c) (c).begin(), (c).end()
#define REVERSE(c) reverse(ALL(c))
#define SORT(c) sort(ALL(c))
#define RSORT(c) sort((c).rbegin(),(c).rend())
#define STABLE_SORT(c) stable_sort(ALL(c))
#define STABLE_RSORT(c) stable_sort((c).rbegin(),(c).rend())
#define UNIQUE(c) SORT(c); (c).erase(unique(ALL(c)), (c).end());
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
const double INF = 1e7;
const double EPS = 1e-10;
const double PI  = acos(-1.0);
#define dump(x)  cerr << #x << " = " << (x) << endl;


int main() {
    LL N, M;
    cin >> N >> M;
    VVI G(N + 1);
    REP(i, M) {
        LL x, y, z;
        cin >> x >> y >> z;
        G[x].PB(y);
        G[y].PB(x);
    }
    vector<PII> S(N);
    FOR(i, 1, N + 1) {
        S.PB(MP(SZ(G[i]), i));
    }
    RSORT(S);
    
    set<LL> covered;
    //sort(G.begin(), G.end(), [](const vector<LL> & a, const vector<LL> & b){ return a.size() > b.size(); });

    /*REP(i, N) {
        cout << "c" << S[i].second << " w size " << S[i].first << endl;
        }*/

    // Choose node with max edges repeatedly
    LL index = 0;
    LL ans = 0;
    while (index < N) {
        if (covered.count(S[index].second) != 0) {
            index++;
            continue;
        }
        LL node = S[index].second;
        covered.insert(node);
        //dump(node);
        for (auto v : G[node]) {
            covered.insert(v);
        }
        index++;
        ans++;
    }
    cout << ans << endl;
	
    return 0;
}