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: Wrong Answer)

import javax.swing.*;
import javax.swing.table.*;
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();

        int[][] T = new int[N][2];
        for (int i = 0; i < N; i++) {
            T[i][0] = in.nextInt();
            T[i][1] = in.nextInt();
        }

        HashMap<BitSet, Long> map = new HashMap<>();
        map.put(new BitSet(), 0L);
        for (int i = M; 0 < i; i--) {
            HashMap<BitSet, Long> tmp = new HashMap<>();

            for (Map.Entry<BitSet, Long> entry : map.entrySet()) {
                BitSet done = entry.getKey();
                Long v = entry.getValue();

                tmp.put(done,v);
                for (int j = 0; j < N; j++) {
                    if (done.get(j)) continue;
                    if (i < T[j][0]) continue;

                    BitSet nk = new BitSet();
                    nk.or(done);
                    nk.set(j);
                    tmp.merge(nk, v + T[j][1], Math::max);
                }
            }
            System.out.println(tmp);

            map = tmp;
        }

        out.println(map.values().stream().max(Comparator.naturalOrder()).orElse(0L));
    }
}