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: Wrong Answer)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let m: i32 = input();
    let mut ab: Vec<(i32, i32)> = (0..n).map(|_| (input(), input())).collect();
    ab.sort_by_key(|&(a, b)| (-a, -b));

    let mut w = std::collections::BinaryHeap::new();
    let mut t = 0;
    let mut ans = 0;
    for (a, b) in ab {
        if t + a <= m {
            t += 1;
            ans += b;
            w.push(Reverse(b));
        } else {
            if let Some(Reverse(x)) = w.pop() {
                if b > x {
                    ans += b - x;
                    w.push(Reverse(b));
                }
            }
        }
    }

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

#[derive(Clone, Copy, Debug, Default, Eq, Hash, PartialEq)]
struct Reverse<T>(T);
impl<T: Ord> Ord for Reverse<T> {
    fn cmp(&self, other: &Reverse<T>) -> std::cmp::Ordering {
        other.0.cmp(&self.0)
    }
}
impl<T: PartialOrd> PartialOrd for Reverse<T> {
    fn partial_cmp(&self, other: &Reverse<T>) -> Option<std::cmp::Ordering> {
        other.0.partial_cmp(&self.0)
    }
}