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: Time Limit Exceeded)

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        // 入力
        int n = sc.nextInt();
        int m = sc.nextInt();

        int[] h = new int[n];
        for (int i = 0; i < n; i++) {
            h[i] = sc.nextInt();
        }

        int[] a = new int[m];
        int[] b = new int[m];
        for (int i = 0; i < m; i++) {
            a[i] = sc.nextInt();
            b[i] = sc.nextInt();
        }

        int[] maxNeighbor = new int[n];
        int currentH;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (a[j] - 1 == i) {
                    // A側が一致する場合
                    currentH = h[b[j] - 1];
                    if (currentH > maxNeighbor[i]) {
                        maxNeighbor[i] = currentH;
                    }
                }
                else if (b[j] - 1 == i) {
                    // B側が一致する場合
                    currentH = h[a[j] - 1];
                    if (currentH > maxNeighbor[i]) {
                        maxNeighbor[i] = currentH;
                    }
                }
            }
        }

        int count = 0;
        for (int i = 0; i < n; i++) {
            if (maxNeighbor[i] == 0 || h[i] > maxNeighbor[i]) {
                count++;
            }
        }

        // 出力
        System.out.println(count);
    }
}