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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
use std::collections::BTreeSet;

fn main() {
    input! {
        n: usize,
        m: usize,
        mut job: [(usize, usize); n],
    }

    // dp[i] = i日間の最大報酬
    let mut dp = vec![0; m + 1];
    // dp[i] = dp[i - 1] + (選べる最大報酬仕事)
    // 初日に追加と考える


    // メモ: 条件は 日数がi - 1以下 の中で 報酬が最大のもの
    //
    // 処理前: 日数が短い順, 報酬が多い順 (日数ごとに分ける？)
    // 処理後: 報酬が多い順(set)

    job.sort_by(|(_, b1), (_, b2)| b1.cmp(b2));
    // eprintln!("{:?}", job);

    // 0は番兵
    let mut future = vec![vec![0]; 100000 + 1];
    for (a, b) in job {
        future[a].push(b);
    }

    // (報酬, 仕事番号)
    let mut processsed = BTreeSet::new();
    // 番兵
    processsed.insert((0, 0));
    let mut job_count = 1;
    for i in 1..=m {
        let (pd_max, pd_i) = *processsed.iter().rev().next().unwrap();
        let future_max = future[i].pop().unwrap();

        if pd_max < future_max {
            dp[i] = dp[i - 1] + future_max;
            for x in &future[i - 1] {
                processsed.insert((*x, job_count));
                job_count += 1;
            }
        } else {
            dp[i] = dp[i - 1] + pd_max;
            processsed.remove(&(pd_max, pd_i));
            processsed.insert((future_max, job_count));
            job_count += 1;
            for x in &future[i] {
                processsed.insert((*x, job_count));
                job_count += 1;
            }
        }
    }

    eprintln!("{:?}", dp);

    let ans = dp[m];
    println!("{}", ans);
}
*/

fn main() {
    let exe = "/tmp/binA218638B";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";