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.util.Map.Entry;
import java.util.*;

import org.w3c.dom.css.Counter;


public class Main {
    public static void main(final String[] args){
        final Scanner s = new Scanner(System.in);
        final int N = Integer.parseInt(s.next());
        final int M = Integer.parseInt(s.next());
        HashSet<Integer> peak = new HashSet<Integer>();
        HashMap<Integer,Integer> nameAndHigh = new HashMap<Integer,Integer>();
        HashSet<Integer> outList = new HashSet<Integer>();
        
        for(int i=1; i <= N ;i++){
           //peak.add(Integer.parseInt(s.next()));
           nameAndHigh.put(i,Integer.parseInt(s.next()));
        }
        
        for(int i=0; i<M ;i++){
            int peakNameOne = Integer.parseInt(s.next());
            int peakNameTwo = Integer.parseInt(s.next());
            int peakOne = nameAndHigh.get(peakNameOne);
            int peakTwo = nameAndHigh.get(peakNameTwo);
            int losePeak;
            if(peakOne<peakTwo){
                losePeak= peakNameOne;
            }else{
                losePeak = peakNameTwo;
            }

            //int hoge = Math.min(Integer.parseInt(s.next()),Integer.parseInt(s.next()));
            if(!outList.contains(losePeak)){
                outList.add(losePeak);
            }
         }

         System.out.println(N-outList.size());;
   

    }
}