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: Accepted)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        int m = std.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];

        for (int i = 0; i < n; i++) {
            a[i] = std.nextInt();
            b[i] = std.nextInt();
        }

        List<List<Integer>> jobs = new ArrayList<>(m);
        for (int i = 0; i < m; i++) {
            jobs.add(new ArrayList<>());
        }

        for (int i = 0; i < n; i++) {
            int v = a[i];
            if (v > m) {
                continue;
            }
            jobs.get(m - v).add(b[i]);
        }

        long ans = 0;
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        for (int i = m - 1; i >= 0; i--) {
            pq.addAll(jobs.get(i));

            if (!pq.isEmpty()) {
                ans += pq.poll();
            }
        }

        System.out.println(ans);
    }
}