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)

/*
 * FileName:     C_fix
 * CreatedDate:  2020-05-04 02:05:14 +0900
 * LastModified: 2020-05-04 02:30:20 +0900
 */

#include <stdio.h>
#include <stdlib.h>
int matrix[100000][100000]={0};
int main(void){
    int n,m;
    scanf("%d",&n);
    scanf("%d",&m);
    int *h = (int *)calloc(n,sizeof(int));
    for(int i=0;i<n;i++){
        scanf("%d",&h[i]);
    }
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d",&a);
        scanf("%d",&b);
        matrix[a-1][b-1]=1;
        matrix[b-1][a-1]=1;
    }
    int flag=0,count=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(matrix[i][j]==1&&h[i]<=h[j]){
                flag=1;
                break;
            }
        }
        if(flag==0){
            count++;
        }
        flag=0;
    }
    printf("%d\n",count);
    free(h);
    return 0;
}