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: Time Limit Exceeded)

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

int main (void)
{
    unsigned long h[100000], *G[100000];
    long i, j, n, m, a, b, ans = 0;

    for (i = 0; i < 100000; i++) {
        G[i] = (unsigned long *)malloc(sizeof(unsigned long) * 100000);
        memset(G[i], 0, sizeof(unsigned long) * 100000);
    }

    scanf("%lo %lo", &n, &m);

    for (i = 0; i < n; i++) {
        scanf("%lu", &(h[i]));
    }

    for (i = 0; i < m; i++) {
        scanf("%lo %lo", &a, &b);
        G[a - 1][b - 1] = 1;
        G[b - 1][a - 1] = 1;
    }

    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            if (G[i][j] == 0) continue;
            if (h[i] <= h[j]) break;
        }
        if (j == n) {
            ans++;
        }
    }

    printf("%lo", ans);

    return 0;
}