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)

#[allow(dead_code)]
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()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}

fn main() {
    let a = read_vec::<usize>();
    let n = a[0];
    let m = a[1];

    let mut v = vec![];
    for _ in 0..n {
        let u = read_vec::<usize>();
        v.push((u[0], u[1] as u64, false));
    }

    v.sort_by_key(|&(_, m, _)| m);

    let mut ans = 0;
    for j in 0..m+1 {
        let mut max = 0;
        let mut maxi = -1;
        for i in (0..n).rev() {
            if !v[i].2 && v[i].0 <= j && v[i].1 > max {
                max = v[i].1;
                maxi = i as i32;
                break;
            }
        }
        //println!("max: {}, j={}, vec={:?}", max, j, v);
        if maxi >= 0 {
            v[maxi as usize].2 = true;
            ans += max;
        }
    }
    println!("{}", ans);
}