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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    public static void main(String[] args) throws NumberFormatException, IOException {
        BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
        int[] first = Arrays.stream(stdin.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
        int n = first[0];
        int m = first[1];
        
        UnionFind uf = new UnionFind(n);
        for (int i = 0; i < m; i++) {
            int[] line = Arrays.stream(stdin.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
            int x = line[0] - 1;
            int y = line[1] - 1;
            int z = line[2];
            
            uf.union(x, y);
        }
        
        int ans = IntStream.range(0, n).map(uf::find).boxed().collect(Collectors.toSet()).size();
        System.out.println(ans);
    }
    
    private static class UnionFind {
        private int[] parents;

        public UnionFind(int n) {
            parents = IntStream.range(0, n).toArray();
        }
        
        public int find(int x) {
            if (parents[x] == x) {
                return x;
            } else {
                parents[x] = find(parents[x]);
                return parents[x];
            }
        }
        
        public void union(int x, int y) {
            x = find(x);
            y = find(y);
            if (x != y) {
                parents[y] = x; 
            }
        }
    }
}
