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: Wrong Answer)

use proconio::input;
 
fn main() {
  input! {
    n: usize,
    m: usize,
    mut abs: [(usize,usize);n]
  }
  abs.sort_by(|a, b| a.1.cmp(&b.1));
  abs.reverse();
  
  let mut result = 0;
  let mut index = m;
  let mut abs_index = 0;
  while 0 < index {
    while abs_index < n {
      let (day, val) = abs[abs_index];
      abs_index += 1; 
      if day <= index {
        result += val;
        break
      }
    }
    index -= 1;
  }
  
  println!("{}", result);
}