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.

Examples

Input

3 1
1 2 1


Output

2


Input

6 5
1 2 1
2 3 2
1 3 3
4 5 4
5 6 5


Output

2


Input

100000 1
1 100000 100


Output

99999
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> ii;
typedef long long ll;

int mark[100005];
vector<int> adj[100005];

void solve(int u){
    mark[u] = 1;
    for(int v : adj[u])
        if(!mark[v])
            solve(v);
}

int main(){
    int n, m; cin >> n >> m;
    int res = 0;
    for(int i = 0; i < m; i++){
        int a, b, c; cin >> a >> b >> c;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    for(int i = 1; i <= n; i++)
        if(!mark[i])
            solve(i),
            res++;
    cout << res << endl;
	return 0;
}