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

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] first = br.readLine().split(" ", 2);
        int n = Integer.parseInt(first[0]);
        int m = Integer.parseInt(first[1]);
        UnionFindTree uft = new UnionFindTree(n);
        for (int i = 0; i < m; i++) {
            String[] line = br.readLine().split(" ", 3);
            int a = Integer.parseInt(line[0]) - 1;
            int b = Integer.parseInt(line[1]) - 1;
            uft.unite(a, b);
        }
        System.out.println(uft.getCount());
    }
    
    static class UnionFindTree {
        int[] parents;

        public UnionFindTree(int size) {
            parents = new int[size];
            for (int i = 0; i < size; i++) {
                parents[i] = i;
            }
        }
        
        public int find(int x) {
            if (parents[x] == x) {
                return x;
            } else {
                return parents[x] = find(parents[x]);
            }
        }
        
        public void unite(int x, int y) {
            int xx = find(x);
            int yy = find(y);
            if (xx == yy) {
                return;
            } 
            parents[xx] = yy;
        }
        
        public int getCount() {
            HashSet<Integer> set = new HashSet<>();
            for (int i = 0; i < parents.length; i++) {
                set.add(find(i));
            }
            return set.size();
        }
    }
}