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: Accepted)

use std::collections::BinaryHeap;

fn read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let n: usize = read(&mut stdin_lock, &mut buf, b' ');
    let m: usize = read(&mut stdin_lock, &mut buf, b'\n');
    let mut abv = vec![(0i32, 0i32); n];
    for i in 0..n {
        abv[i].0 = read(&mut stdin_lock, &mut buf, b' ');
        abv[i].1 = read(&mut stdin_lock, &mut buf, b'\n');
    }

    let mut min_a_heap = BinaryHeap::new();
    for i in 0..n {
        min_a_heap.push((m as i32 - abv[i].0, abv[i].1));
    }
    let mut max_b_heap = BinaryHeap::new();
    let mut ans = 0;
    for i in 0..m {
        loop {
            match min_a_heap.peek() {
                None => break,
                Some(&(ra, b)) => {
                    if m as i32 - ra > (i + 1) as i32 {
                        break;
                    }
                    min_a_heap.pop();
                    max_b_heap.push(b);
                }
            }
        }
        match max_b_heap.pop() {
            None => continue,
            Some(b) => ans += b,
        }
    }
    println!("{}", ans);
}