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: Wrong Answer)

import java.io.FileNotFoundException;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws FileNotFoundException {

//    	File file = new File("src/in.txt");
//    	Scanner sc = new Scanner(file);

    	Scanner sc = new Scanner(System.in);
    	int N = sc.nextInt();
    	int M = sc.nextInt();
    	int[] H = new int[N];
    	int[] A = new int[M];
    	int[] B = new int[M];
    	for(int i=0;i<N;i++) {
    		H[i] = sc.nextInt();
    	}
    	for(int i=0;i<M;i++) {
    		A[i] = sc.nextInt();
    		B[i] = sc.nextInt();
    	}
    	int ans = 0;
    	for(int i=0;i<N;i++) {
    		boolean flg = true;
    		int h = H[i];
    		//Aに入っている場合
    		for(int j=0;j<M;j++) {
    			if(A[j]==i) {
    				int h2 = H[B[j]-1];
    				if(h2>h) {
    					flg = false;
    				}
    			}
    		}
    		for(int j=0;j<M;j++) {
    			if(B[j]==i) {
    				int h2 = H[A[j]-1];
    				if(h2>h) {
    					flg = false;
    				}
    			}
    		}
    		if(flg) {
    			ans ++;
    		}

    	}
    	System.out.println(ans);
    }
}