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.*;

public class Main {
    private static int[] roots;
    
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        UnionFind uf = new UnionFind(n);
        
        while (m-- > 0) {
            int x = sc.nextInt() - 1;
            int y = sc.nextInt() - 1;
            sc.nextInt();
            uf.unite(x, y);
        }
        
        System.out.println(uf.connectedComponent);
    }
    
    public static class UnionFind {
        int[] par;
        int connectedComponent;
    
        UnionFind(int n) {
            par = new int[n];
            connectedComponent = n;
            for (int i = 0; i < n; i++) {
                par[i] = i;
            }
        }
    
        int root(int x) {
            if (par[x] == x) return x;
            par[x] = root(par[x]);
            return par[x];
            // 1-2-4の場合、4の親=2を4の根=1に繋ぎ変えた上で親を返す
            // ↑根まで辿る深さを減らす為
        }
    
        void unite(int x, int y) {
            int rx = root(x);
            int ry = root(y);
            if (rx != ry) {
                par[ry] = rx;
                connectedComponent--;
            }
        }
    
        boolean same(int x, int y) {
            int rx = root(x);
            int ry = root(y);
            return rx == ry;
        }
        
        void printPar() {
            System.out.println(Arrays.toString(par));
        }
    }
}