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>
#include <stdlib.h>
#include <string.h>

int maxind(int *H);

int M;
int N;
int *H;
int *A;
int *B;
int *T;

int main(void)
{
    int c;
    int ans = 0;
    int max;
    int temp;
    int start;
    scanf("%d", &N);
    scanf("%d", &M);
    H = (int *)malloc(sizeof(int) * N);
    A = (int *)malloc(sizeof(int) * M);
    B = (int *)malloc(sizeof(int) * M);
    T = (int *)malloc(sizeof(int) * N);
    for (c = 0; c < N; ++c)
    {
        scanf("%d", H + c);
        T[c] = c;
    }
    for (c = 0; c < M; ++c)
    {
        scanf("%d", A + c);
        scanf("%d", B + c);
        --A[c];
        --B[c];
    }
    max = maxind(H);
    start = H[max];
    while (start > 0)
    {
        H[max] *= -1;
        temp = 1;
        for (c = 0; c < M; ++c)
        {
            if (A[c] == max)
            {
                if (abs(H[B[c]]) == start)
                {
                    temp = 0;
                }
                if (H[B[c]] > 0)
                {
                    H[B[c]] *= -1;
                }
            }
            else if (B[c] == max)
            {
                if (abs(H[A[c]]) == start)
                {
                    temp = 0;
                }
                if (H[A[c]] > 0)
                {
                    H[A[c]] *= -1;
                }
            }
        }
        ans += temp;
        max = maxind(H);
        start = H[max];
    }
    printf("%d", ans);
    free(A);
    free(B);
    free(H);
    free(T);
    return 0;
}

int maxind(int *H)
{
    int temp = H[0];
    int ans = 0;
    int c;
    for (c = 1; c < N; ++c)
    {
        if (H[c] > temp)
        {
            temp = H[c];
            ans = c;
        }
    }
    return ans;
}