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)

use std::cmp::{max, min};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::io::{stdin, Read};
use std::str::FromStr;

// https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

#[allow(dead_code)]
fn main() {
    let n = read::<usize>();
    let m = read::<usize>();
    let ab = (0..n)
        .map(|_| (read::<usize>(), read::<usize>()))
        .collect::<Vec<_>>();
    let mut t = vec![vec![]; m + 1];

    for (a, b) in ab {
        if a > m {
            continue;
        }
        t[a].push(b);
    }
    for i in 1..m {
        t[i].sort()
    }
    let mut ans = 0;

    let mut x = BinaryHeap::new();
    for i in 1..m + 1 {
        for j in 0..t[i].len() {
            x.push(t[i][j]);
        }

        let v = x.pop();
        if v.is_none() {
            continue;
        } else {
            ans += v.unwrap();
        }
    }
    println!("{}", ans);
}