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 main(void){
	int n, m;
	long *h;
	int *a;
	int *b;
	int *flgs;
	scanf("%d", &n);
	scanf("%d", &m);
	h = (long *)malloc(sizeof(long) * n);
	flgs = (int *)malloc(sizeof(int) * n);
	a = (int *)malloc(sizeof(int) * m);
	b = (int *)malloc(sizeof(int) * m);
	for(int i=0; i<n; i++){
		scanf("%ld", &h[i]);
		flgs[i] = 1;
	}
	for(int j=0; j<m; j++){
		scanf("%d", &a[j]);
		scanf("%d", &b[j]);
	}
	for(int k=0; k<m; k++){
		int tmp_a = a[k] - 1;
		int tmp_b = b[k] - 1;
		if(h[tmp_a] > h[tmp_b]){
			flgs[tmp_b] = 0;
		}
		else if(h[tmp_a] < h[tmp_b]){
			flgs[tmp_a] = 0;
		}
		else{
			flgs[tmp_a] = 0;
			flgs[tmp_b] = 0;
		}
	}
	int sum = 0;
	for(int l=0; l<n; l++){
		if(flgs[l] == 1){
			sum++;
		}
	}
	printf("%d\n", sum);
	return 0;
}