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: Accepted)

<?php

fscanf(STDIN, "%d %d", $n, $m);
$A = [];
for ($i = 0; $i < $n; ++$i) {
    fscanf(STDIN, "%d %d", $a, $b);
    $A[$m - $a][] = $b;
}

$sum = 0;
$pq = new SplPriorityQueue();
for ($i = $m - 1; $i >= 0; --$i) {
    // i日目にできる仕事は$A[$i]
    if (isset($A[$i])) {
        foreach ($A[$i] as $val) {
            $pq->insert($val, $val);
        }
    }

    if ($pq->valid()) {
        $sum += $pq->extract();
    }
}

echo $sum;