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


int main() {
	int i, j, k;
	int m, n;
	int f = 0;
	int g = 0;
	int h[110000];
	int a[110000], b[110000];
	int c[110000] = { 0 };

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

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

	for (i = 0;i < m;i++) {
		scanf("%d %d", &a[i], &b[i]);
		c[a[i]]++;
		c[b[i]]++;
	}
	
	for (i = 0;i < n;i++) {
		g = 0;
		if (c[i + 1] == 0) {
			goto fin;
		}
		for (j = 0;j < m;j++) {
			if (a[j] == (i + 1)) {
				if (h[i] <= h[(b[j] - 1)]) {
					g = 1;
					break;
				}
			}else if (b[j] == (i + 1)) {
				if (h[i] <= h[(a[j] - 1)]) {
					g = 1;
					break;
				}
			}
		}
		fin:
		if (g == 0) {
			f++;
		}
	}

	printf("%d", f);
	

	

	return 0;
}