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(void)
{
    int n,m;
    scanf("%d %d",&n,&m);
    // printf("%d %d\n",n,m);

    long long int h[n];
    for (int i=0;i<n;i++){
        scanf("%ld",&h[i]);
        // printf("%d\n",h[i]);
    }

    long long int a[m];
    long long int b[m];
    long long int c[m];
    long long int d[m];
    for (int i=0;i<m;i++){
        scanf("%ld %ld",&a[i],&b[i]);
        // printf("%d %d\n",a[i],b[i]);
        if (h[a[i]-1]<h[b[i]-1]){
            // printf("%d\n",h[a[i]-1]);
            c[i]=h[a[i]-1];
            d[i]=0;
        }else if (h[a[i]-1]>h[b[i]-1]){
            // printf("%d\n",h[b[i]-1]);
            c[i]=h[b[i]-1];
            d[i]=0;
        }else{ 
            c[i]=h[a[i]-1];
            d[i]=h[b[i]-1];
        }
        // printf("c[%d]=%d\n",i,c[i]);
    }

    for (int j=0;j<n;j++){
        for (int i=0;i<m;i++){
            // printf("h[%d]=%d\n",j,h[j]);
            if (h[j]==c[i])
                h[j]=0;
            if (h[j]==d[i])
                h[j]=0;
            // printf("c[%d]=%d\n",i,c[i]);
            // printf("d[%d]=%d\n",i,d[i]);
            // printf("h[%d]=%d\n",j,h[j]);
        }
    }

    int ans=0;
    for (int j=0;j<n;j++){
        if (h[j]!=0)
        ans=ans+1;
    }

    printf("%d\n",ans);

    return 0;


}