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 <malloc.h>
#include <string.h>

int main()
{
	int n, m;
	int* h/*標高*/, * a, * b/*展望台を結ぶ道*/;
	scanf("%d %d", &n, &m);
#pragma region input
	h = (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("%d", h + i);
	}
	for (int i = 0; i != m; i++)
	{
		scanf("%d %d", a + i, b + i);
	}
#pragma endregion

	short* s = (short*)malloc(n * sizeof(short));
	memset(s, 1, sizeof(short));

	for (int i = 0; i != m; i++)
	{
		if (*(h + (*(a + i))) > * (h + (*(b + i))))
		{
			*(s + (*(b + i) - 1)) = 0;
		}
		else
		{
			*(s + (*(a + i) - 1)) = 0;
		}
	}

	int z = 0;
	for (int i = 1; i != n; i++)
	{
		if (*(s + i))
		{
			z++;
		}
	}

	printf("%d", z);

	return 0;
}