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: Accepted)

#include <stdio.h>
 

int main(){
  int n = 2, m = 1;
  int a[ 100001 ] = {0};
  int a_c[ 100001 ] = {0};

  scanf( "%d %d", &n, &m );
  
  int i = 1;
  for ( i = 1; i < n + 1; i++ ){
    int h = 0;
    scanf( "%d", &h );
    a[ i ] = h;
    a_c[ i ] = 1;
  }

  
  for ( i = 1; i < m + 1; i++ ){
    int ai = 0, bi = 0;
    scanf( "%d %d", &ai, &bi );
    if ( a[ ai ] < a[ bi ] ){
      a_c[ ai ] = 0;
    }else if ( a[ ai ] > a[ bi ] ){
      a_c[ bi ] = 0;
    }else if ( a[ ai ] == a[ bi ] ){
      a_c[ ai ] = 0;
      a_c[ bi ] = 0;
    }
  }
  
  int c = 0;
  for ( i = 1; i < n + 1; i ++ ){
    c += a_c[ i ];
  }

  printf("%d\n", c);

  return 0;
  
}