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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(){
	int n, m, i, j, count;
	scanf("%d %d", &n, &m);
	int h[n], a[m], b[m], counter[n], no[n];
	for(i=0; i<n; i++){
		scanf("%d", &h[i]);
		counter[i]=0;
		no[i]=0;
	}
	for(i=0; i<m; i++){
		scanf("%d %d", &a[i], &b[i]);
		if(a[i]<=b[i]){
			no[a[i]-1]=1;
		}else if(a[i]>=b[i]){
			no[b[i]-1]=1;
		}
		//no[a[i]-1]=1;
		//no[b[i]-1]=1;
	}
	/*for(i=0; i<m; i++){
		if(h[a[i]]>h[b[i]]){
			counter[a[i]-1]=1;
			counter[b[i]-1]=0;
		}else if(h[a[i]]<h[b[i]]){
			counter[b[i]-1]=1;
			counter[a[i]-1]=0;
		}
		if(h[a[i]]==h[b[i]]){
			counter[b[i]-1]=0;
			counter[a[i]-1]=0;
		}
	}*/
	count=0;
	for(i=0; i<n; i++){
		if(/*(counter[i]==1)||(*/no[i]==0){
			count++;
		}
	}
	printf("%d", count);

	return 0;
}