Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
fscanf(STDIN, "%d %d", $n, $m);

$job = [];

for(; $n>0; $n--){
  fscanf(STDIN, "%d %d", $a, $b);

  $job[] = [
    'day' => $a,
    'money' => $b,
  ];
}

$money = 0;
for($i=$m; $i>0; $i--){
  $job = array_filter($job, function($v) use($i) {return $v['day'] <= $i;});

  if(count($job) === 0) break;

  $key = selectjob($job, $i);
  $money += $job[$key]['money'];
  unset($job[$key]);
}

echo $money, "\n";

function selectjob($jobs, $day) {
  $tmp = [];

  foreach($jobs as $k => $job){
    $w = $job['money'] + ($job['day'] - $day);
    $tmp[$w] = $k;
  }

  ksort($tmp);

  return array_pop($tmp);
}