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 java.util.Collections;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		sc.nextInt();
		Long m = sc.nextLong();
		TreeMap<Long, Queue<Long>> map = new TreeMap<>();
		while (sc.hasNextLong()) {
			Long a = sc.nextLong(); // a日後
			Long b = sc.nextLong(); // 報酬
			Queue<Long> list = map.getOrDefault(a, new PriorityQueue<Long>(Collections.reverseOrder()));
			list.add(b);
			map.put(a, list);
		}

		// 日数ごとにバイトの報酬順でのソート

		long sum = 0;
		for (long i = m - 1; i >= 0; i--) {
			Long maxKye = -1L;
			Long maxValue = -1L;
			for (Long key : map.keySet()) {
				if (m - i >= key) {
					Queue<Long> temp = map.get(key);
					if (temp.isEmpty()) {
						continue;
					}
					maxValue = Math.max(maxValue, temp.peek());
					maxKye = key;
				}
			}
			if (maxKye < 0 && maxValue < 0) {
				break;
			}
			sum = maxValue + sum;
			Queue<Long> temp = map.get(maxKye);
			if (temp.isEmpty()) {
				break;
			}
			temp.poll();
		}
		sc.close();
		System.out.println(sum);
	}
}