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>

int	main(void)
{
  	long long	n;
  	long long 	m;
  	long long	h[100002];
  	long long	a[100002];
  	long long	b[100002];
  	long long	i;
  	long long 	j;
  	long long   flag;
  	long long   cnt;
  	long long   flag2;
  
  	scanf("%lld %lld\n", &n, &m);
  	for (i=0; i<n; i++)
    {
      	scanf("%lld", &h[i]);
    }
  	scanf("\n");
  
    for (j=0; j<m; j++)
    {
      	scanf("%lld %lld\n", &a[j], &b[j]);
    }
    
    cnt = 0;
    flag = 0;
    flag2 = 0;
	for(i=0; i<n; i++)
	{
	    for (j=0; j<m; j++)
	    {
	        if (i == a[j])
	        {
	            if (h[i] <= h[ b[j] ])
	            {
	                continue ;
	            }
	            else
	            {
	                flag = 1;
	            }
	        }
	        else if (i == b[j])
	        {
	            if (h[i] <= h[ a[j] ])
	            {
	                continue ;
	            }
	            else
	            {
	                flag = 1;
	            }
	        }
	        else
	        {
	        }
	    }
	    if (flag == 1)
	    {
	        cnt++;
	        flag = 0;
	    }
	}
	printf("%lld", cnt);
	return (0);
}