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: Time Limit Exceeded)

macro_rules! get {
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
                    get!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
                    get!($($t),*)
        ).collect::<Vec<_>>()
    };
    ($t:ty |) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty | $n:expr) => {
        (0..$n).map(|_| get!($t |)).collect::<Vec<_>>()
    };
}
macro_rules! tern {
    ($condition: expr, $_true: expr, $_false: expr) => {
        if $condition {
            $_true
        } else {
            $_false
        }
    };
}

fn main() {
    let (N, M) = get!(usize, usize);
    let mut ABs = get!(usize, usize; N);
    use std::collections::BinaryHeap;
    let mut heaps = (0..(10001)).map(|b| {
        BinaryHeap::new()
    }).collect::<Vec<_>>();
    let mut max_B = 0;
    let mut min_B = 10000;
    for (A, B) in ABs {
        if B > max_B {
            max_B = B;
        }
        if B < min_B {
            min_B = B;
        }
        heaps[B].push(-(A as i64));
    }
    let mut sum = 0;
    for d in 1..(M+1) {
        for b in (min_B..(max_B+1)).rev() {
            let mut flg = false;
            if let Some(arev) = heaps[b].peek() {
                let min_a = -arev;
                if min_a as usize <= d {
                    flg = true;
                    sum += b;
                }
            } else {
                if b == max_B {
                    max_B -= 1;
                }
            }
            if flg {
                heaps[b].pop();
                break;
            }
        }
    }
    println!("{}", sum);
}