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
import java.util.*; 

class Main{
public static void main(String[] args){
	Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
		
  		int[] h=new int[n];
		for(int c=0;c<n;c++){
    		h[c]=sc.nextInt();
    	}
  		int[] c=new int[n];
  		int[] a=new int[m];
  		int[] b=new int[m];
  		for(int d=0;d<m;d++){
    		a[d]=sc.nextInt();
          	b[d]=sc.nextInt();
          if(h[a[d]-1]<h[b[d]-1]){
          c[a[d]-1]++;
          }else if(h[a[d]-1]==h[b[d]-1]){
           c[a[d]-1]++;
           c[b[d]-1]++;
          }else{
          c[b[d]-1]++;
          }
    	}
  	int ans=0;
    for(int e=0;e<n;e++){
    if(c[e]==0){
    ans++;
    }
    }
 	

    	System.out.println(ans);
    }
}