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 1.8.0) (judge verdict: Runtime Error)

import java.math.BigInteger;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		long M = Integer.parseInt(sc.next());
		long H[] = new long[N];
		long Way[][] = new long[N][N];
		int p, q;
		int i, j;
		for (i = 0; i < N; i++) {
			H[i] = Integer.parseInt(sc.next());
		}
		for (i = 0; i < M; i++) {
			p = Integer.parseInt(sc.next());
			q = Integer.parseInt(sc.next());
			Way[p - 1][q - 1] = 1;
			Way[q - 1][p - 1] = 1;
		}
		//ここまでinput
		long count = 0;
		boolean flag = true;
		BigInteger high = BigInteger.valueOf(1);
		for (i = 0; i < N; i++) {
			high = BigInteger.valueOf(H[i]);
			flag = true;
			j = 0;
			while (flag && j < N) {
				if (Way[i][j] > 0 && (high.compareTo(BigInteger.valueOf(H[j])) <= 0)) {
					//					System.out.println(
					//							(i + 1) + "展望台は" + high + "で" + (j + 1) + "展望台は" + H[j] + "だから" + (j + 1) + "展望台の方が高い");
					break;
				} else if (j == N - 1) {
					flag = false;
				} else {
					j++;
				}

			}
			if (j == N - 1 && flag == false) {
				count++;
			}
		}
		System.out.println(count);
		sc.close();
	}
}