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.*;
class Main{
    public static void main(String[] args){
        // ----------入力-------------//
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();           // 展望台の数
        int M = sc.nextInt();           // 道の数

        // ---展望台の高さを記憶---//
        int[] height = new int[N];      // 展望台の高さ
        for(int i=0;i<N;i++){
            height[i] = sc.nextInt();
        }
        //-------良い展望台を識別するフラグ good を用意--------//
        boolean[] good = new boolean[N];    //良い展望台のときtrue
        Arrays.fill(good, true);            // trueで初期化
        show(good);

        // ---道が入力されるたびに比較を行い，低い方はgoodフラグを下ろす---//
        int a,b;                // 道がつなぐ二つの展望台
        for(int j=0;j<M;j++){
            a = sc.nextInt() -1;   // 一つ目の展望台
            b = sc.nextInt() -1;   // 二つ目の展望台
            if(height[a]==height[b]){
                // 何もしない
            }
            else if(height[a]<height[b]){   // aが低い時，aのフラグを下ろす
                good[a]=false;
            }
            else{                               // bが低い時，bのフラグを下ろす
                good[b]=false;
            }
            show(good);
        }
        // ---------出力-----------//
        int cnt=0;
        for(int k=0;k<N;k++){
            if(good[k]){
                cnt++;
            }
        }
        System.out.println();
        System.out.println(cnt);
    }

    static void show(boolean[] array){
        for(int l=0;l<array.length;l++){
            System.out.println(array[l]);
        }
    }
}