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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.*;
import java.util.*;

public class Main {
    public static final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    public static final PrintStream out = System.out;

    public static void main(String[] args) {
        int N = in.nextInt();
        int M = in.nextInt();

        NavigableMap<Integer, List<Integer>> T = new TreeMap<>();
        for (int i = 0; i < N; i++) {
            T.computeIfAbsent(in.nextInt(), k -> new LinkedList<>())
                    .add(in.nextInt());
        }
        T.forEach((k, v) -> v.sort(Comparator.reverseOrder()));

        int cost = 0;
        for (int i = 1; i <= M; i++) {
            Map.Entry<Integer, List<Integer>> maxE = null;
            for (Map.Entry<Integer, List<Integer>> entry : T.headMap(i, true).entrySet()) {
                if (entry.getValue().isEmpty()) {
                    continue;
                }
                
                if (maxE == null) {
                    maxE = entry;
                } else if (maxE.getValue().get(0) < entry.getValue().get(0)) {
                    maxE = entry;
                }
            }

            if (maxE != null) {
                cost += maxE.getValue().remove((int) 0);
            }
        }

        out.println(cost);
    }
}