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.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[][] ab = new int[n][2];
		for (int i = 0; i < n; i++) {
			ab[i][0] = sc.nextInt();
			ab[i][1] = sc.nextInt();
		}

		Arrays.sort(ab, Comparator.comparing(a -> a[0]));
		long a = 0;
		int idx = 0;

		PriorityQueue<Integer> pq = new PriorityQueue<>(Comparator.reverseOrder());
		for (int i = 1; i <= m; i++) {
			while (idx < ab.length && ab[idx][0] <= i) {
				pq.add(ab[idx++][1]);
			}
			if (!pq.isEmpty()) a += pq.poll();
		}

		System.out.println(a);
	}
}