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::{HashSet, HashMap, BinaryHeap};

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    buf.trim().split_whitespace()
        .map(|x| x.parse().ok().unwrap()).collect()
}

fn main() {
    let nm = read_vec::<i64>();
    let (n, m) = (nm[0], nm[1]);
    let mut AB = Vec::<(i64, i64)>::with_capacity(n as usize);

    for i in 0..n as usize {
        let ab = read_vec::<i64>();
        AB.push((ab[0], ab[1]));
    }
    AB.sort();
    let mut res : i64 = 0;
    let mut idx : usize = 0;
    let mut heap = BinaryHeap::new();
    for i in 1..m+1 {
        while idx < AB.len() && AB[idx].0 <= i {
            heap.push(AB[idx].1);
            idx += 1;
        }
        if heap.len() > 0 {
            res += heap.pop().unwrap();
        }
    }
    println!("{}", res);
}