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>

int main(void)
{
    long n,m,h[100000],ab[100000][2],i,j,con[100001][100001]={0},
    ans=0,ifgood;

    scanf("%ld %ld",&n,&m);
    for ( i = 0; i < n; i++)
    {
        scanf("%ld",h+i);
    }
    for ( i = 0; i < m; i++)
    {
        scanf("%ld %ld",&ab[i][0],&ab[i][1]);
        long a=ab[i][0];
      	long b=ab[i][1];
        con[a][0]++;
        con[b][0]++;
        con[a][con[a][0]]=b;
        con[b][con[b][0]]=a;
    }

    for ( i = 1; i <= n; i++)
    {
        ifgood=1;
        j=1;
        while(j<=100000&&con[i][j]!=0)
        {
            if(h[con[i][j]]>=h[i]) ifgood=0;
            j++;
        }
        if(ifgood=1) ans++;
    }

    printf("%ld",ans);

    return 0;
}