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.*;
import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int ans = 0;
		int n = scan.nextInt();
		int m = scan.nextInt();
		int h[] = new int[n];
		int a[][]= new int[m][2];
		int non[] = new int[n];
		
		for(int i=0;i<n;i++) {
			h[i] = scan.nextInt();
		}
		
		for(int i=0;i<m;i++) {
			for(int j=0;j<2;j++) {
				a[i][j] = scan.nextInt();
			}
			
			if(h[a[i][0]-1]<h[a[i][1]-1]) {
				non[a[i][0]-1] = 1;
			}else if(h[a[i][0]-1]>h[a[i][1]-1]){
				non[a[i][1]-1] = 1;
			}else if(h[a[i][0]-1]==h[a[i][1]-1]){
				non[a[i][1]-1] = 1;
				non[a[i][0]-1] = 1;
			}
		}
		
		for(int i=0;i<n;i++) {
			if(non[i] == 0) {
				ans += 1;
			}
		}
		
		System.out.println(ans);
	}	
	
	
	//最小公約数
	/*
	static int gcd (int a, int b) {
		int temp;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return b;
	}
	*/
	
}