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.ArrayList;
import java.util.*;
 class Main
{
	public static void main (String[] args)
	 {
		Scanner sc=new Scanner(System.in); 
		int i,j;
		int n=sc.nextInt();
		int m=sc.nextInt();
		HashMap<Integer,Integer> h=new HashMap();
		HashMap<Integer,Integer> v=new HashMap();
		for(i=1;i<=n;i++)
		{
			h.put(i, sc.nextInt());
			v.put(i,1);
		}
		
		int in,og;
		for(i=0;i<m;i++)
		{
			in=sc.nextInt();
			og=sc.nextInt();
			
		 if(h.get(in)>=h.get(og))
			{
				v.put(og, -1);
			}
			if(h.get(in)<=h.get(og))
			{
				v.put(in,-1);
			}
		}
		int count=0;
		for(i=1;i<=n;i++)
		{
			if(v.get(i)!=-1)
				count++;
		}
		
		
			
		System.out.println(count);
		
			
	}
}