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: Accepted)

<?php
$arr = explode(' ', trim(fgets(STDIN)));

$n = $arr[0];
$m = $arr[1];

$h_arr = explode(' ', trim(fgets(STDIN)));

for ($i=1; $i <= $n; $i++) { 
    $h_nums[] = $i;
}

for ($i=0; $i < $m; $i++) { 
    $arr2 = explode(' ', trim(fgets(STDIN)));
    $a = $arr2[0];
    $b = $arr2[1];

    // echo "anum =".$a.PHP_EOL;
    // echo "bnum =".$b.PHP_EOL;
    // echo "a =".$h_arr[$a - 1].PHP_EOL;
    // echo "b =".$h_arr[$b - 1].PHP_EOL;

    if($h_arr[$a - 1] > $h_arr[$b - 1]) {
        $ng_nums[] = $b;
    } elseif($h_arr[$a - 1] == $h_arr[$b - 1]) {
        $ng_nums[] = $a;
        $ng_nums[] = $b;
    } else
    {
        $ng_nums[] = $a;
    }
}
$ng_nums = array_unique($ng_nums);
// for ($i=0; $i < count($ng_nums); $i++) { 
//     echo "ng_nums = ".$ng_nums[$i].PHP_EOL;
// }

$h_nums = array_diff($h_nums, $ng_nums);
$h_nums = array_values($h_nums);
// for ($i=0; $i < count($h_nums); $i++) { 
//     echo "h_nums = ". $h_nums[$i].PHP_EOL;
// }

echo count($h_nums);

return;
?>