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: Runtime Error)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Objects;
import java.util.List;
import java.util.Scanner;
import java.util.Comparator;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Peaks solver = new Peaks();
        solver.solve(1, in, out);
        out.close();
    }

    static class Peaks {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();

            int[] h = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                h[i] = in.nextInt();
            }

            int[][] loads = new int[n + 1][n + 1];

            List<List<Pair>> list = new ArrayList<>();
            for (int i = 0; i <= n; i++) {
                list.add(new ArrayList<>());
            }

            for (int i = 1; i <= m; i++) {
                int a = in.nextInt();
                int b = in.nextInt();

                (list.get(a)).add(new Pair(h[b], b));
                (list.get(b)).add(new Pair(h[a], a));
            }

            boolean[] result = new boolean[n + 1];
            for (int i = 1; i <= n; i++) {
                List<Pair> directions = list.get(i);
                if (directions.isEmpty()) {
                    result[i] = true;
                    continue;
                }
                directions.sort(Comparator.reverseOrder());
                Pair top = directions.get(0);
                if (h[top.dist] > h[i]) {
                    result[top.dist] = true;
                } else if (h[top.dist] < h[i]) {
                    result[i] = true;
                }

            }

            int count = 0;
            for (int i = 1; i <= n; i++) {
                if (result[i]) {
                    count++;
                }
            }
            out.append(String.valueOf(count));
        }

        public class Pair implements Comparable<Peaks.Pair> {
            public final Integer cost;
            public final Integer dist;

            public Pair(Integer cost, Integer dist) {
                this.cost = cost;
                this.dist = dist;
            }

            public boolean equals(Object o) {
                if (this == o) return true;
                if (o == null || getClass() != o.getClass()) return false;
                Pair pair = (Pair) o;
                return cost.equals(pair.cost) &&
                        dist.equals(pair.dist);
            }

            public int hashCode() {
                return Objects.hash(cost, dist);
            }

            public int compareTo(Pair o) {
                return Integer.compare(cost, o.cost);
            }

        }

    }
}