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 larger(int a,int b){
 if(a<b)return b;
 else   return a;
}

int main(){
 long long  n,m;
 scanf("%lld%lld",&n,&m);
 long long  h[n],a[m],b[m];
 long long i;
 for(i=0;i<n;i++)
  scanf("%lld",&h[i]);
 
 for(i=0;i<n;i++)
  scanf("%lld%lld",&a[i],&b[i]);
  
 long long  cnt[n],tmp[n],x;
 for(i=0;i<n;i++){
  cnt[i]=0;tmp[i]=0;}
  
 for(i=0;i<m;i++){
  cnt[a[i]-1]++;cnt[b[i]-1]++;
  if(h[a[i]-1]>h[b[i]-1])tmp[a[i]-1]++;
  if(h[a[i]-1]<h[b[i]-1])tmp[b[i]-1]++;
 }
 
 long long ans=0;
 for(i=0;i<n;i++){
 // printf("%lld %lld %lld\n",i+1,cnt[i],tmp[i]);
  if(cnt[i]==tmp[i])ans++;}
  
 printf("%lld\n",ans);
  
return 0;}