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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int main(void){
    int n, m;
    int *h;
    int *a;
    int *is_good;
    int i;
    int result = 0;

    scanf("%d %d", &n, &m);
    h = (int*)malloc(sizeof(int) * n);
    is_good = (int*)malloc(sizeof(int) * n);
    a = (int*)malloc(sizeof(int) * m * 2);

    for(i = 0; i < n; i++){
        scanf("%d", h + i);
        is_good[i] = 1;
    }
    for(i = 0; i < m; i++){
        scanf("%d %d", a + (i * 2), a + (i * 2 + 1));
    }

    for(i = 0; i < m; i++){
        int h1 = h[a[i * 2] - 1];
        int h2 = h[a[i * 2 + 1] - 1];
        if(h1 <= h2)
            is_good[a[i * 2] - 1] = 0;
        else if(h1 >= h2)
            is_good[a[i * 2 + 1] - 1] = 0;
    }

    for(i = 0; i < n; i++){
        if(is_good[i] == 1)
            result++;
    }



    free(is_good);
    free(a);
    free(h);
    printf("%d\n", result);
    return 0;
}