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::collections::BinaryHeap;

fn solve(abs: Vec<(usize, usize)>, m: usize) -> usize {
    let mut m = m;
    let mut abs = abs;

    abs.sort_by_key(|&(a, b)| a);

    let mut ans = 0;

    let mut heap = BinaryHeap::new();

    let mut idx = 0;
    for i in 1..(m + 1) {
        while idx < abs.len() && abs[idx].0 == i {
            heap.push(abs[idx].1);
            idx += 1;
        }

        if let Some(max) = heap.pop() {
            ans += max;
        }
    }

    ans
}

#[test]
fn solve_test() {
    assert_eq!(solve(vec![(4, 3), (4, 1), (2, 2)], 4), 5);
    assert_eq!(solve(vec![(1, 2), (1, 3), (1, 4), (2, 1), (2, 3)], 3), 10);
    assert_eq!(solve(vec![(2, 1)], 1), 0);
}

#[allow(unused_macros)]
macro_rules! parse_line {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            iter.next().unwrap().parse::<$t>().unwrap()
        }
    );

    ( $( $t:ty), +) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ( $(iter.next().unwrap().parse::<$t>().unwrap()),* )
        }
    );
}

#[allow(unused_macros)]
macro_rules! read_line {
    () => {{
        let mut line = String::new();
        ::std::io::stdin().read_line(&mut line).unwrap();
        line.pop();
        line
    }};
}

#[allow(unused_macros)]
macro_rules! parse_vec {
    ( $t:ty ) => {{
        let mut line = String::new();
        ::std::io::stdin().read_line(&mut line).unwrap();
        let iter = line.split_whitespace();
        iter.map(|v| v.parse::<$t>().unwrap()).collect::<Vec<_>>()
    }};
}

fn main() {
    let (n, m) = parse_line!(usize, usize);

    let abs = (0..n)
        .map(|_| parse_line!(usize, usize))
        .collect::<Vec<_>>();
    println!("{}", solve(abs, m));
}