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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		UnionFind uf = new UnionFind(n);
		int m = sc.nextInt();
		for (int i = 0; i < m; i++) {
			int x = Integer.parseInt(sc.next())-1;
			int y = Integer.parseInt(sc.next())-1;
			uf.unite(x, y);
			sc.next();
		}
		System.out.println(uf.count());
		sc.close();
	}

	public static class UnionFind {

		int par[]; int rank[]; int size[]; int count;

		public UnionFind(int n) {
			this.par = new int[n];
			this.rank = new int[n];
			this.size = new int[n];
			this.count = n;
			for(int i = 0; i < n; i++) {
				par[i] = i; rank[i] = 0; size[i] = 1;
			}
		}

		int find(int x) {//xの属する集合の根を探す
			if(par[x] == x) return x;
			else return par[x] = find(par[x]);
		}

		void unite(int x, int y) {//x,yの集合を合併
			x = find(x);
			y = find(y);
			if(x == y) return ;
			if(rank[x] < rank[y]) {
				par[x] = y;
				size[y] += size[x];
			}else {
				par[y] = x;
				size[x] = size[y];
				if(rank[x] == rank[y]) {
					rank[x]++;
				}
			}
			count--;
		}

		boolean same(int x, int y) {//x,yが同じ集合に属するか
			return find(x) == find(y);
		}

		int amount(int x) {//xの属する集合の要素数
			return size[find(x)];
		}

		int count() {//集合の数
			return count;
		}
	}
}