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


int main(){

 long long int N,M;
  scanf("%lld %lld",&N,&M);

/*
  char A[];
  scanf("%s",A);
  */

  
  long long int H[N+1];
  long long int i;
  for(i=1;i<=N;i++){
   scanf("%lld ",&H[i]);
   }
  
  double T[N+1];
    for (i=1;i<=N;i++){
      T[N]=1;
    }
  long long int A[M+1],B[M+1];
  for(i=1;i<=M;i++){
   scanf("%lld %lld",&A[i],&B[i]);
   if (H[A[i]] < H[B[i]]){
     if(T[B[i]]!=-1){
     T[A[i]] = -1;
     T[B[i]] = 1;
     }
     else{
     T[A[i]] = -1;
     }
   }
   else if (H[A[i]] > H[B[i]]){
     if(T[A[i]]!=-1){
     T[B[i]] = -1;
     T[A[i]] = 1;
     }
     else{
     T[B[i]] = -1;
     }

   }
   else{
     T[A[i]] = -1;
     T[B[i]] = -1;

   }


   }

long long int counter;
counter = 0;
for (i=1;i<=N;i++){
  if(T[i]==1){
    counter +=1;
  }
}
printf("%lld",counter);






 return(0);
 }