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;
use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    let mut it = s.split_whitespace();
    let (n, m) = parse!(it, 2);
    let mut items: Vec<(usize, usize)> = (0..n).map(|_| parse!(it, 2)).collect();
    items.sort();
    items.reverse();
    let mut q = BinaryHeap::new();
    let mut ret = 0;
    for i in 0..m {
        while items.len() > 0 {
            let last = items[items.len() - 1];
            if last.0 <= i + 1 {
                q.push(last.1);
                items.pop();
            } else {
                break;
            }
        }
        if let Some(v) = q.pop() {
            ret += v;
        }
    }
    println!("{}", ret);
}