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.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int A[] = new int[N];
		int B[] = new int[N];
		for(int i=0;i<N;i++) {
			A[i]=sc.nextInt();
			B[i]=sc.nextInt();
		}
		
		Queue<Integer> pq = new PriorityQueue<Integer>((i1,i2)->i2-i1);
		
		int C[][] =  new int[N][2];
		for(int i=0;i<N;i++) {
			C[i][0]=A[i];
			C[i][1]=B[i];
		}
		Arrays.sort(C,(c1,c2)->c1[0]-c2[0]);//Aの昇順に並んでいる
		
		long ans = 0;
		int j=0;
		for(int i=1;i<=M;i++) {
			while(j<N && C[j][0]<=i) {
				pq.offer(C[j][1]);
				j++;
			}
			if(!pq.isEmpty()) ans += pq.poll();
		}
		System.out.println(ans);
	}
}