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: Time Limit Exceeded)

<?php
$sum = [0 => PHP_INT_MAX];
$ans = 0;
fscanf(STDIN, "%d %d", $N, $M);
$arr = array_fill(1, $M, []);
for ($i = 0; $i < $N; $i++) {
    fscanf(STDIN, "%d %d", $A, $B);
    $arr[$A][] = $B;
}

for ($i = 1; $i <= $M; $i++) {
    foreach ($arr[$i] as $dummy => $v) {
        if (!isset($sum[$v])) {
            $sum[$v] = 1;
        } else {
            $sum[$v]++;
        }
    }

    $mmax = max(array_keys($sum));
    $ans += $mmax;
    if ($sum[$mmax] === 1) {
        unset($sum[$mmax]);
    } else {
        $sum[$mmax]--;
    }
}
echo $ans;