Problem: There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N. The elevation of Obs. i is H_i. There are also M roads, each connecting two different observatories. Road j connects Obs. A_j and Obs. B_j.

Obs. i is said to be good when its elevation is higher than those of all observatories that can be reached from Obs. i using just one road. Note that Obs. i is also good when no observatory can be reached from Obs. i using just one road.

How many good observatories are there?

Constraints

* 2 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq H_i \leq 10^9
* 1 \leq A_i,B_i \leq N
* A_i \neq B_i
* Multiple roads may connect the same pair of observatories.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
H_1 H_2 ... H_N
A_1 B_1
A_2 B_2
:
A_M B_M


Output

Print the number of good observatories.

Examples

Input

4 3
1 2 3 4
1 3
2 3
2 4


Output

2


Input

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6


Output

3

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
		int[] height;		
		height = new int[n];		//標高
		for (int i=0; i<n; i++){
			height[i] =  scan.nextInt();
		}

		HashSet<Integer> hs = new HashSet<>();
		for (int i=0; i<n; i++){
			hs.add(i);
		}

		int a = -1;
		int b = -1;
		for (int i=0; i<m; i++){
			a = scan.nextInt()-1;
			b = scan.nextInt()-1;
//			System.out.println(a +":"+ b);

			if(height[a]>height[b]){
				hs.remove(b);
			} else if(height[a]<height[b]) {
				hs.remove(a);
			} else {		//同じ高さの場合は両方「良い展望台」に該当せず
				hs.remove(a);
				hs.remove(b);
			}
		}
		
		System.out.println(hs.size());
	    scan.close();	
    }
}