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.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        Main m = new Main(sc);
        m.show(m.solve());
        sc.close();
    }
    
	//フィールド
    Scanner sc;
    int N;
    int M;
    int[] H;
    int[] A;
    int[] B;
    
    //入力
    Main(Scanner sc) {
        this.sc = sc;
        this.N = sc.nextInt();
        this.M = sc.nextInt();
        this.H = new int[N];
        this.A = new int[N];
        this.B = new int[N];
        for (int i = 0; i < N; i++) {
        	this.H[i] = sc.nextInt();
        }
        for (int i = 0; i < M; i++) {
        	this.A[i] = sc.nextInt();
        	this.B[i] = sc.nextInt();
        }
    }
    
    //解答処理
    private int solve() {
        int count = 0;
        int[] m = new int[N];
        
        for (int i = 0; i < M; i++) {
        	m[A[i]-1] = Math.max(m[A[i]-1], H[B[i]-1]);
        	m[B[i]-1] = Math.max(m[B[i]-1], H[A[i]]-1);
        }
        
        for (int i = 0; i < N; i++) {
        	if (m[i] < H[i]) {
        		count++;
        	}
        }
        
        return count;
    }
    
    //出力
    public void show(int answer) {
        System.out.println(answer);
    }

}