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);
		int n = sc.nextInt();
		int m = sc.nextInt();
		// 4 3
		// 1 2 3 4
		// 1 3
		// 2 3
		// 2 4
		Map<Integer, PriorityQueue<Integer>> graph = new HashMap<>();
		for (int i = 0; i < n; i++) {
			graph.put(i + 1, new PriorityQueue<>(m, Collections.reverseOrder()));
		}
		int[] heights = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			heights[i] = sc.nextInt();
		}
		for (int i = 0; i < m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			graph.get(a).add(heights[b]);
			graph.get(b).add(heights[a]);
		}
		int ans = 0;
		for (Map.Entry<Integer, PriorityQueue<Integer>> entry : graph.entrySet()) {
			int node = entry.getKey();
			PriorityQueue<Integer> maxHeap = entry.getValue();
			if (maxHeap.isEmpty() || maxHeap.peek() < heights[node]) {
				ans++;
			}
		}
		System.out.println(ans);
	}
}