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.*;
public 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];
    int b[]=new int[n];
    int c=0;
    for(int i=0;i<n;i++)
    {
      h[i]=sc.nextInt();
     b[i]=0;
    }
    int x,z;
    for(int i=0;i<m;i++)
    {
      x=sc.nextInt();
      z=sc.nextInt();
      b[x-1]=Math.max(b[x-1],h[z-1]);
        b[z-1]=Math.max(b[z-1],h[x-1]);
    }
    
     for(int i=0;i<n;i++)
     {
       if(h[i]>b[i])
       {
        // System.out.println(i+1);
         c++;
       }
         
       
     }
    
    
     System.out.println(c);
    
  }
}