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)

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

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

use std::collections::BinaryHeap;

fn main() {
    let nm: Vec<usize> = read_vec();
    let n = nm[0];
    let m = nm[1] as u32;
    let mut ab: Vec<Vec<u32>> = read_vec2(n);
    let mut bh: BinaryHeap<u32> = BinaryHeap::new();
    ab.sort_by(|x, y| {
        if x[0] == y[0] {
            y[1].partial_cmp(&x[1]).unwrap()
        } else {
            x[0].partial_cmp(&y[0]).unwrap()
        }
    });

    let mut ans: u32 = 0;

    let mut j: usize = 0;
    for i in 1..(m + 1) {
        while j < ab.len() {
            if ab[j][0] <= i {
                bh.push(ab[j][1]);
                j += 1;
            } else {
                break;
            }
        }
        if let Some(x) = bh.pop() {
            ans += x;
        }
    }
    println!("{}", ans);
}