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>

#define MAX_N 200001
#define max(a,b) ((a) > (b) ? (a):(b))
#define min(a,b) ((a) < (b) ? (a):(b))

typedef struct{
  int number;
  int high;
}tower;

int main(void){

  int N,M,H[MAX_N],A[MAX_N],B[MAX_N],i,j,mem[MAX_N] = {0},ans = 0,big,small,big_n,small_n;
  tower ViewTower[MAX_N];

  scanf("%d %d",&N,&M);
  for(i = 1;i <= N;i++){
    scanf("%d",&ViewTower[i].high);
    ViewTower[i].number = i;
  }
  for(i = 1;i <= M;i++){
    scanf("%d %d",&A[i],&B[i]);
  }

  for(i = 1;i <= M;i++){
    big = max(ViewTower[A[i]].high,ViewTower[B[i]].high);
    small = min(ViewTower[A[i]].high,ViewTower[B[i]].high);

    if(big == ViewTower[A[i]].high){
      big_n = ViewTower[A[i]].number;
      small_n = ViewTower[B[i]].number;
    }else{
      big_n = ViewTower[B[i]].number;
      small_n = ViewTower[A[i]].number;
    }
    
    //printf("big:%d small:%d big_n:%d small_n:%d\n",big,small,big_n,small_n);
    if(mem[big_n] >= 0) mem[big_n]++;
    mem[small_n] = -1;
  }

  for(i = 1;i <= N;i++){
    //printf("mem[%d]:%d\n",i,mem[i]);
    if(mem[i] >= 0) ans++;
  }

  printf("%d\n",ans);
  return 0;
}