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 sc = new Scanner(System.in);
      
      int n = sc.nextInt();
      int m = sc.nextInt();
      
      int a[] = new int[n];
      
      int b[] = new int[n];
      
      for (int g = 0; g < n; g++) {
         a[g] = sc.nextInt();
      }
      
      for (int h = 0; h < n; h++) {
         b[h] = 0;
      }
      
      for (int i = 0; i < m; i++) {
         int x = sc.nextInt();
         int y = sc.nextInt();
         if (a[x-1] > a[y-1]) {
            b[y-1] = 1;
         } else if (a[x-1] == a[y-1]) {
            b[x-1] = 1;
            b[y-1] = 1;
         } else {
            b[x-1] = 1;
         }
      }
      
      int count = 0;
      
      for (int l = 0; l < n; l++) {
         if (b[l] == 0) {
            count++;
         }
      }
      
      System.out.println(count);
   }
}