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::vec::Vec;
use std::collections::BinaryHeap;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let (n, m) = {
      let v: Vec<usize> = read_vec();
      (v[0], v[1])
    };

    // (b_i, a_i)
    let mut v: Vec<(u64, u64)> = Vec::new();
    for _ in 0..n {
      let t: Vec<u64> = read_vec();
      v.push((t[0], t[1]));
    }
    v.sort_by(|a, b| b.0.cmp(&a.0));
    //println!("{:?}", v);
    let mut heap = BinaryHeap::new();
    let mut now = v.pop().unwrap();
    let mut ans = 0;
    for i in 1..m + 1 {
      while now.0 <= i as u64 {
        heap.push(now.1);
        now = match v.pop() {
          None => (1_000_000, 0),
          Some(e) => e,
        };
      }
      ans += match heap.pop() {
        None => 0,
        Some(x) => x,
      };
    }

    //println!("{:?}", heap);
    println!("{}", ans);

}