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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner(System.in);
		long n = sc.nextLong(), m = sc.nextLong();
		long[] list = new long[(int)n];
		for(int i = 0; i < n; i++) {
			list[i] = sc.nextLong();
		}
		Map<Integer,Integer> map = new HashMap<>();
		ArrayList<Integer> li = new ArrayList<>(); 
		for(int i = 0; i < m; i++) {
			int a = sc.nextInt(), b = sc.nextInt();
			map.put(a, 1);
			map.put(b, 1);
			
			if(list[a-1] > list[b-1]) {
				if(!li.contains(a))li.add(a);
			}else if(list[a-1] < list[b-1]) {
				if(!li.contains(b))li.add(b);
			}
		}
		long ans = n-map.size()+li.size();
		System.out.println(ans);
	}

	static class FastScanner {
		private BufferedReader reader = null;
	    private StringTokenizer tokenizer = null;
	    public FastScanner(InputStream in) {
	        reader = new BufferedReader(new InputStreamReader(in));
	        tokenizer = null;
	    }

	    public String next() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }

	    public String nextLine() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                return reader.readLine();
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }

	        return tokenizer.nextToken("\n");
	    }

	    public long nextLong() {
	        return Long.parseLong(next());
	    }

	    public int nextInt() {
	        return Integer.parseInt(next());
	    }

	    public double nextDouble() {
	         return Double.parseDouble(next());
	     }

	    public int[] nextIntArray(int n) {
	        int[] a = new int[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextInt();
	        return a;
	    }

	    public long[] nextLongArray(int n) {
	        long[] a = new long[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextLong();
	        return a;
	    }
	}
}