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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php

fscanf(STDIN, "%d %d", $in_n, $in_m);
$in_h = trim(fgets(STDIN));
$h = explode(" ", $in_h);
//var_dump($h);
$ans = 0;
$miti = array_fill(0, $in_n, array());


for($i = 0; $i < $in_m; $i++) {
    fscanf(STDIN, "%d %d", $in_a[$i], $in_b[$i]);
//    var_dump($in_b[$i]);
    array_push($miti[$in_a[$i] - 1], $in_b[$i]);
    array_push($miti[$in_b[$i] - 1], $in_a[$i]);
}


for($i = 0; $i < $in_n; $i++) {
//    var_dump($miti[$i]);
    $flag = true;
    for ($j = 0; $j < count($miti[$i]); $j++) {
        if($h[$i] < $h[$miti[$i][$j]]){
            $flag = false;
            break;
        }
    }
    if(($flag == true) && (count($miti[$i]) > 0)){
        $ans++;
    }
}

echo "$ans\n";