Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

4 3
1 2 3 4
1 3
2 3
2 4

Sample Output 1

2

From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.

From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.

From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.

From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.

Thus, the good observatories are Obs. 3 and 4, so there are two good observatories.

Sample Input 2

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6

Sample Output 2

3

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    resolve(System.in, System.out);
  }

  static void resolve(InputStream is, PrintStream pw) {
    Scanner sc = new Scanner(is);
    int n = Integer.parseInt(sc.next());
    int m = Integer.parseInt(sc.next());
    Node[] nodes = new Node[n + 1];
    for (int i = 0; i < n; i++) {
      nodes[i + 1] = new Node(i + 1, Integer.parseInt(sc.next()));
    }
    for (int j = 0; j < m; j++) {
      int a = Integer.parseInt(sc.next());
      int b = Integer.parseInt(sc.next());
      nodes[a].addOther(nodes[b].h);
      nodes[b].addOther(nodes[a].h);
    }
    int ans = 0;
    for (int i = 1; i < n + 1; i++) {
      if (nodes[i].isGood()) {
        ans++;
      }
    }
    pw.println(ans);
  }

  static class Node {

    int i;
    int h;
    int otherH = 0;

    public Node(int i, int h) {
      this.i = i;
      this.h = h;
    }

    void addOther(int otherH) {
      this.otherH = Math.max(this.otherH, otherH);
    }

    boolean isGood() {
      return h > otherH;
    }
  }
}