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)

fn read_line() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim_right().to_owned()
}

fn read_stdin() -> String {
    use std::io::Read;
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    s
}

fn main() {
    let nums: Vec<u64> = read_line().split_whitespace().map(|x| x.parse().unwrap()).collect();
    let (_, m) = (nums[0] as usize, nums[1]);
    let mut jobs: Vec<Vec<u64>> = read_stdin().lines()
        .map(|line: &str| line.split_whitespace().map(|s| s.parse().unwrap()).collect())
        .collect();
    jobs.sort();
    jobs = jobs.into_iter().rev().collect();
    
    let mut total = 0;
    for day in (0..m).rev() {
        if let Some(i) = jobs.iter().position(|j| day + 1 >= j[0]) {
            total += jobs[i][1] as u64;
            jobs.remove(i);
        }
    }
    
    println!("{}", total);
}