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>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define BUFSIZE 10000000
typedef long long ll;
#define WDSIZE 10000

ll root[WDSIZE][WDSIZE]={0};
ll a_root[BUFSIZE]={0};
ll b_root[BUFSIZE]={0};
int count = 0;
ll mount_high[BUFSIZE]={0};
int N,M;

int cal(int mount_num){
  if(mount_num == N + 1) return 0;
  for(int i = 1;i < N + 1;i++){
       if(root[mount_num][i] == 1 || root[i][mount_num] == 1){
          if(mount_high[i] >= mount_high[mount_num]){
             break;
           }
          }
        if(i == N){

          count++;
        }
    }
    cal(mount_num + 1);
}


int main(void){


  scanf("%d %d",&N,&M);
  for(int i = 1;i <= N;i++){
    scanf("%lld",&mount_high[i]);
    }

  for(int i = 0;i < M; i++){
      scanf("%lld",&a_root[i]);
      scanf("%lld",&b_root[i]);
      root[a_root[i]][b_root[i]] = 1;

    }

    cal(1);
    printf("%d",count);

}