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.util.*;

public class Main {
    public static void main(String[] args) {
        try (Scanner s = new Scanner(System.in)) {
            int n = s.nextInt();
            int m = s.nextInt();
            s.nextLine();
            int[] h = Arrays.stream(s.nextLine().split(" "))
                            .mapToInt(Integer::parseInt)
                            .toArray();
            var way = new HashMap<Integer, Set<Integer>>();
            for (int i = 0; i < m; i++){
                int a = s.nextInt() - 1, b = s.nextInt() - 1;
                way.computeIfAbsent(a, k -> new HashSet<>()).add(b);
                way.computeIfAbsent(b, k -> new HashSet<>()).add(a);
            }
            long good = (n - way.size()) + way.entrySet().stream()
                .filter(e -> e.getValue().stream().allMatch(v -> h[e.getKey()] > h[v]))
                .count();
            System.out.println(good);
        }
    }
}