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.IOException;
import java.io.InputStream;
import java.util.HashSet;
import java.util.NoSuchElementException;

public class Main {
    private static FastScanner sc = new FastScanner();
    private static boolean DEBUG_FLG = false;
    static Node[] V;

    public static void main(String[] args) {
    	int N = sc.nextInt();
    	int M = sc.nextInt();
    	V = new Node[N];
    	for(int i=0; i<N; i++) {
    		V[i] = new Node();
    	}
    	for(int i=0; i<M; i++) {
    		int X = sc.nextInt() - 1;
    		int Y = sc.nextInt() - 1;
    		int Z = sc.nextInt();
    		V[X].nbr.add(Y);
    		V[Y].nbr.add(X);
    	}
    	int ans = 0;
    	for(int i=0; i<N; i++) {
    		if(!V[i].vst) {
    			V[i].dfs();
        		ans ++;
    		}
    	}
    	System.out.println(ans);
    }

    static class Node {
        HashSet<Integer> nbr = new HashSet<>();
        boolean vst = false;
        void dfs() {
        	vst = true;
        	for(int x : nbr) {
        		if(!V[x].vst) {
        			V[x].dfs();
        		}
        	}
        }
    }

    static void debug(long... args) {
    	if(!DEBUG_FLG) return;
    	boolean flg = false;
    	for(long s : args) {
    		if(flg) System.out.print(" ");
    		flg = true;
    		System.out.print(s);
    	}
    	System.out.println();
    }

    static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte() {
            if(ptr < buflen) {
                return true;
            } else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch(IOException e) {
                    e.printStackTrace();
                }
                if(buflen <= 0) {
                    return false;
                }
            }
            return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
        public boolean hasNext() { skipUnprintable(); return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
    }
}
