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: Runtime Error)

#include<stdio.h>
int main(void){
	int n,m,count=0;
	long long int h[100000];
	scanf("%d %d",&n,&m);
	for(int i=0;i<n;i++){
		scanf("%lld",&h[i]);
	}
	int road[n][n+1];
	for(int i=0;i<n;i++){
		for(int j=0;j<n+1;j++){
			road[i][j]=0;
		}
	}
	int temp1,temp2;
	for(int i=0;i<m;i++){
		scanf("%d %d",&temp1,&temp2);
		road[temp1-1][temp2-1]=1;
		road[temp2-1][temp1-1]=1;
	}
	for(int i=0;i<n;i++){
		if(road[i][n]==0){
			temp1=0;
			for(int j=i+1;j<n;j++){
				if(road[i][j]==1){
					if(h[i]>h[j]){
						road[j][n]=1;
					}else if(h[i]<h[j]){
						temp1=1;
					}else{
						road[j][n]=1;
						temp1=1;
					}
				}
			}
		}else{
			temp1=1;
		}
		//printf("%d",temp1);
		if(temp1==0){
			count++;
		}
	}
	printf("%d",count);
}