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

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

$n = $a[0];
$days = $a[1];

$tmps = [];
for ($i = 0; $i < $n; $i++) {
    $tmps[] = explode(' ', trim(fgets(STDIN)));
}

$arr = [];

for ($i = 0; $i < $n; $i++) {
    $arr[$i]['day'] = $tmps[$i][0];
    $arr[$i]['money'] = $tmps[$i][1];
}

$day = array_column($arr, 'day');
$money = array_column($arr, 'money');

array_multisort($money, SORT_DESC, $day, SORT_DESC, $arr);


$ans = 0;
while (count($arr) !== 0) {
    // 先頭から取っていく
    $record = array_shift($arr);
    if ($record['day'] <= $days) {
        $ans += $record['money'];
        $days -= 1;
    }
}

echo $ans;