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: Time Limit Exceeded)

#include <stdio.h>

int main(){
  	int ans = 0;
  	int i, j;
  	int N, M;
  	int A, B;
  	int NextNum;
  	int High[100000];
  	int Road[100000][2];
  	int NextTower[100000];
  	int BadTower[100000];
  
  	scanf("%d%d", &N, &M);
  	//input - High
  	for(i=0;i<N;i++){
      scanf("%d", &High[i]);
    }
  	//input - Road
  	for(j=0;j<M;j++){
    	scanf("%d%d", &Road[j][0], &Road[j][1]);
    }
  
  	//init
  	for(i=0;i<N;i++){
    	BadTower[i] = 0;
    }
  	
  	//check
  	for(i=0;i<N;i++){
      if(BadTower[i] == 1){
        //printf("skip:%d\n", i);
         continue;
      }
      for(j=0;j<N;j++){
        NextTower[j] = 0;
      }
      NextNum = 0;
      for(j=0;j<M;j++){
        A = Road[j][0]-1;
        B = Road[j][1]-1;
       if(A == i){
          	NextTower[NextNum++] = B;
          	if(High[B] >= High[i]){
              break;
            }
        }else if(B == i){
          	NextTower[NextNum++] = A;
        	if(High[A] >= High[i]){
              break;
            }
        }else{
        	//This tower has no next tower.
        }
      }
      if(j==M){
        ans++;
        for(j=0;j<NextNum;j++){
        	BadTower[NextTower[j]] = 1;
        }
      }
    }
  
  	printf("%d\n", ans);
  	
	return 0;
}