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.InputStreamReader;
import java.util.Scanner;

public class Main {
	public static void main(final String[] args) {
		@SuppressWarnings("resource")
		final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		final int N = in.nextInt();
		final int M = in.nextInt();
		final int X[] = new int[M];
		final int Y[] = new int[M];
		final int Z[] = new int[M];
		for (int i = 0; i < M; ++i) {
			X[i] = in.nextInt();
			Y[i] = in.nextInt();
			Z[i] = in.nextInt();
		}
		int cost = solve(N, M, X, Y, Z);
		System.out.println(cost);
	}

	public static class UnionFind {
		final int[] parent;
		final int[] rank;
		int disjointCount;

		public UnionFind(final int N) {
			this.parent = new int[N];
			this.rank = new int[N];
			for (int i = 0; i < N; ++i) {
				parent[i] = i;
			}
			this.disjointCount = N;
		}

		public int find(final int i) {
			if (parent[i] == i)
				return i;
			int top = find(parent[i]);
			parent[i] = top;
			return top;
		}

		public void unite(int i, int j) {
			i = find(i);
			j = find(j);
			if (i == j)
				return;
			--disjointCount;
			if (rank[i] < rank[j]) {
				parent[i] = j;
			} else {
				parent[j] = i;
				if (rank[i] == rank[j])
					++rank[i];
			}
		}

		public boolean same(int i, int j) {
			return find(i) == find(j);
		}

		public int disjointCount() {
			return disjointCount;
		}
	}

	public static int solve(int N, int M, int[] X, int[] Y, int[] Z) {
		final UnionFind uf = new UnionFind(N);
		for (int i = 0; i < M; ++i) {
			uf.unite(X[i] - 1, Y[i] - 1);
		}
		return uf.disjointCount();
	}
}
