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: Accepted)

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

int N, M;
int H[114514];

int ec[114514];
int* es[114514];

void ae(int f, int t) {
	es[f] = realloc(es[f], sizeof(*es[f]) * (ec[f] + 1));
	if (es[f] == NULL) exit(2);
	es[f][ec[f]++] = t;
}

int main(void) {
	int i, j;
	int count = 0;
	if (scanf("%d%d", &N, &M) != 2) return 1;
	for (i = 1; i <= N; i++) {
		if (scanf("%d", &H[i]) != 1) return 1;
	}
	for (i = 0; i < M; i++) {
		int A, B;
		if (scanf("%d%d", &A, &B) != 2) return 1;
		ae(A, B);
		ae(B, A);
	}
	for (i = 1; i <= N; i++) {
		int yes = 1;
		for (j = 0; j < ec[i]; j++) {
			if (H[i] <= H[es[i][j]]) yes = 0;
		}
		if (yes) count++;
	}
	printf("%d\n", count);
	return 0;
}