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 (1.15.1) (judge verdict: Time Limit Exceeded)

use std::collections::BinaryHeap;
use std::io;

struct PartTimeJob {
    a: i64,
    b: i64,
}

fn main() {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();

    let mut line_one_iter = buf.split_whitespace();

    let n: i64 = line_one_iter.next().unwrap().parse().unwrap();
    let m: i64 = line_one_iter.next().unwrap().parse().unwrap();

    let mut part_time_jobs_list: Vec<PartTimeJob> = Vec::new();

    for _ in 0..n {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();

        let mut iter = buf.split_whitespace();

        let a: i64 = iter.next().unwrap().parse().unwrap();
        let b: i64 = iter.next().unwrap().parse().unwrap();

        let part_time_job = PartTimeJob { a: a, b: b };

        part_time_jobs_list.push(part_time_job);
    }

    let mut total_salary = 0;
    let mut days_left = 1;
    let mut primary_queue: BinaryHeap<i64> = BinaryHeap::new();

    for _ in 0..m {
        for job in &part_time_jobs_list {
            if job.a <= days_left && job.a > days_left - 1 {
                primary_queue.push(job.b);
            }
        }
        if !primary_queue.is_empty() {
            let salary = primary_queue.pop().unwrap();
            total_salary += salary
        }
        days_left += 1;
    }

    println!("{}", total_salary);
}