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.*;
	import java.util.Map.Entry;
	 
	 
	 class Main {
		 static int mod =  (int) (Math.pow(10,9)+7);
		 static List<ArrayList<Integer>>  list = new ArrayList<ArrayList<Integer>>();
	    public static void main(String[] args) {
	    	
	        Scanner sc = new Scanner(System.in);
	        int N = sc.nextInt();	
	        int M = sc.nextInt();
	        long[] H = new long[N];
	        for(int i=0;i<N;i++) {
	        	H[i] = Long.parseLong(sc.next());
	        }
	        for(long i=0;i<N;i++)list.add(new ArrayList<Integer>());
	        for(int i=0;i<M;i++) {
	        	int a=sc.nextInt()-1;
	        	int b=sc.nextInt()-1;
	        	list.get(a).add(b);
	        	list.get(b).add(a);
	        }
        	
        	
	        long ans=0;
	        for(int i=0;i<N;i++) {
	        	boolean highest=true;
	        	for(int k:list.get(i)) {
	        		if(H[k]>=H[i])highest=false;
	        	}
	        	if(highest==true)ans++;
	        }
	        System.out.println(ans);
	        
	    }
	}