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)

//sieu nhan
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define inp "test.inp"
#define out "test.out"
#define oo 100005
#define inf 1e15
#define mod 1000000007
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define y1 hungdzno1
int n,m,res,x,y,z;
bool ok[oo];
void nhap(){
  cin>>n>>m;
  for(int i=1;i<=m;++i){
    cin>>x>>y>>z;
    if(!ok[x] && !ok[y]){
      ++res;
      ok[x]=1;
      ok[y]=1;
    }
    if(ok[x] || ok[y]){
      ok[x]=1;
      ok[y]=1;
    }
  }
}
void xuly(){
  for(int i=1;i<=n;++i) if(!ok[i]) ++res;
  cout<<res;
}
main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  //freopen(inp,"r",stdin);
  //freopen(out,"w",stdout);
  nhap();
  xuly();
}