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 C (Clang 10.0.0) (judge verdict: Wrong Answer)

//
//  main.c
//  material
//
//  Created by 古屋和樹 on 2020/02/14.
//  Copyright © 2020 古屋和樹. All rights reserved.
//

#include <stdio.h>

int main(int argc, const char * argv[]) {
    int n, m, count = 0;
    scanf("%d%d\n", &n, &m);
    
    long h[n];
    int a[n],b[n];
    
    for(int i = 0; i < n; i++){
        scanf("%ld", &h[i]);
    }
    
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &a[i], &b[i]);
    }
    
    for(int i = 0; i < n; i++){
        for (int j = 0; j < m; j++) {
            if ((a[j] == i-1 && h[i] < h[b[j]-1])|| (b[i] == i-1&& h[i] < h[b[j]-1])) {
                break;
            }else if (j == m-1){
                count++;
            }
        }
    }
    
    printf("%d", count);
    
    return 0;
}