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: Wrong Answer)

import java.util.Scanner;
public class Main{
    public static void main(String[] args){
        int N,M;

        Scanner scn = new Scanner(System.in);
        N = scn.nextInt();
        M = scn.nextInt();
        long[] h = new long[N];
        int[] a = new int[M];
        int[] b = new int[M];

        boolean[] ans = new boolean[N];
        //Hの入力
        for(int i=0;i<N;i++){
            h[i] = scn.nextLong();
        }
        //A,Bの入力と計算
        for(int i=0;i<M;i++){
            a[i] = scn.nextInt();
            b[i] = scn.nextInt();
        }

        int cnt = 0;

        //計算
        for(int i=0;i<M;i++){
            if(h[a[i]-1] <= h[b[i]-1]){
                ans[b[i]-1] = true;
            }else{
                ans[a[i]-1] = false;
            }
        }

        for(int i=0;i<N;i++){
            if(ans[i]){
                //System.out.println("i:"+i);
                cnt++;
            }
        }
        System.out.println(cnt);
    }
}