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>

int main(){
    int N, M;
    scanf("%d %d", &N, &M);

    int H[N];
    int i, j;
    for (i = 0; i < N;i++){
        scanf("%d", &H[i]);
    }

    int A[M];
    int B[M];

    for (i = 0; i < M;i++){
        scanf("%d %d", &A[i], &B[i]);
    }

    int takasa_A[N];
    int takasa_B[N];
    int yoi = 0;
    int check[N];

    for (i = 0; i < N;i++){
        check[i] = 0;
    }

        for (i = 0; i < M; i++)
        {
            takasa_A[i] = H[A[i]];
            takasa_B[i] = H[B[i]];
            if (takasa_A[i] < takasa_B[i])
            {
                if (check[B[i]] == 0){
                    yoi++;
                    check[B[i] - 1] += 1;
                }
                
            }
            else if (takasa_A[i] > takasa_B[i])
            {
                if (check[A[i]] == 0){
                    yoi++;
                    check[A[i] - 1] += 1;
                }
                
            }
        }

    printf("%d", yoi);
    return 0;
}