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)

#![allow(unused_imports)]

use std::cmp::{max, min};
use std::collections::{BTreeMap, BTreeSet, HashMap, HashSet, VecDeque};
use std::io::stdin;
use std::ops::*;
use std::process::exit;
use std::str::FromStr;

fn main() {
    let (n, m): (usize, usize) = input_two();
    let mut ab: Vec<(usize, usize)> = input_lines_two(n);
    let mut done = vec![false; n];

    let mut ans = 0;
    for i in (1..m + 1).rev() {
        let p = ab
            .iter()
            .enumerate()
            .filter(|x| !done[x.0] && (x.1).0 <= i)
            .max_by_key(|x| (x.1).1);

        match p {
            Some((i, &(_, b))) => {
                done[i] = true;
                ans += b;
            }
            None => continue,
        }
    }

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

pub fn input<T: FromStr>() -> T {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().ok().unwrap()
}

pub fn input_vec<T: FromStr>() -> Vec<T> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim()
        .split_whitespace()
        .map(|x| x.parse().ok().unwrap())
        .collect::<Vec<_>>()
}

pub fn input_lines<T: FromStr>(l: usize) -> Vec<T> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input());
    }
    res
}

pub fn input_lines_two<T: FromStr, U: FromStr>(l: usize) -> Vec<(T, U)> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input_two());
    }
    res
}

pub fn input_lines_three<T: FromStr, U: FromStr, V: FromStr>(l: usize) -> Vec<(T, U, V)> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input_three());
    }
    res
}

pub fn input_two<T: FromStr, U: FromStr>() -> (T, U) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse::<T>().ok().unwrap(),
        s[1].parse::<U>().ok().unwrap(),
    )
}

pub fn input_three<T: FromStr, U: FromStr, V: FromStr>() -> (T, U, V) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse::<T>().ok().unwrap(),
        s[1].parse::<U>().ok().unwrap(),
        s[2].parse::<V>().ok().unwrap(),
    )
}

pub fn input_four<T: FromStr, U: FromStr, V: FromStr, W: FromStr>() -> (T, U, V, W) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse().ok().unwrap(),
        s[1].parse().ok().unwrap(),
        s[2].parse().ok().unwrap(),
        s[3].parse().ok().unwrap(),
    )
}