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: Time Limit Exceeded)

import java.util.*;
import static java.lang.Math.*;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
	    HashMap<Integer, PriorityQueue<Integer>> map = new HashMap<Integer, PriorityQueue<Integer>>();
	    for(int i = 0; i < n; i++){
	        int a = sc.nextInt();
	        int b = sc.nextInt();
            if(map.containsKey(a)){
                map.get(a).add(b);
            }else{
                PriorityQueue<Integer> pq = new PriorityQueue<Integer>(Comparator.reverseOrder());
                pq.add(b);
                map.put(a, pq);
            }
		}
		
		// 計算
		int sum = 0;
        for(int i = 1; i < m + 1; i++){
            int max_i = 0;
            int max_val = 0;
            for(int j = 1; j < i + 1; j++){
                if(map.containsKey(j) && map.get(j).peek() != null){
                    if(max_val < map.get(j).peek()){
                        max_val = map.get(j).peek();
                        max_i = j;
                    }
                }
            }
            if(map.containsKey(max_i)) sum = sum + map.get(max_i).poll();
        }

		// 出力
        System.out.println(sum);
		
	}
}