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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
use std::collections::BinaryHeap;
 
fn main() {
  input! {
    n: usize,
    m: usize,
    mut abs: [(usize,usize);n]
  }
  abs.sort_by(|a, b| a.0.cmp(&b.0));
  abs.reverse();
  
  let mut p_queue = BinaryHeap::new();
  let mut result = 0;
  let mut index = m;
  for index in 1..=m {
    while !abs.is_empty() {
      let last = abs.last().unwrap();
      if last.0 > index {
        break
      }
      p_queue.push(abs.pop().unwrap().1);
    }
    
    if let Some(v) = p_queue.pop() {
      result += v;
    }
  }
  
  println!("{}", result);
}