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
import java.io.*;
import java.util.*;


public class Main {
    static int n,count;
    static int[] id;
    public static void main(String[] args) throws Exception{
        //Scanner sc = new Scanner(System.in);
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] buf = reader.readLine().split(" ");
        n = Integer.parseInt(buf[0]);
        int m = Integer.parseInt(buf[1]);
        count = n;
        id = new int[n+1];
        for(int i=0;i<=n;i++) id[i]=i;
        for(int i=0;i<m;i++){
            buf = reader.readLine().split(" ");
            int u = Integer.parseInt(buf[0]), v = Integer.parseInt(buf[1]);
            int len = Integer.parseInt(buf[2]);
            union(u,v);
        }
        System.out.println(count);
    }
    static int find(int p){
        while(id[p]!=p){
            id[p] = id[id[p]];
            p = id[p];
        }
        return p;
    }
    static void union(int p, int q){
        int i = find(p), j = find(q);
        if(i==j) return;
        id[j]=i;
        count--;
    }
}

