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.*;

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();
		}
		int m1 = 1;
		int ans = 0;
		while(m1 != m+1){
			int j = 0,max = 0,limit = 0;
			for(int i = 0; i < n; i++){
				if(b[i] > max){
				max = b[i];
				j = i;
			}
				if(b[i] == max){
					max = b[i];
					if(a[j] < a[i]){
						j = i;
					}
				}
			}
			if(a[j] <= m1){
			    ans += b[j];
				//System.out.println(b[j]);
				b[j] = 0;
			}else{
				break;
			}
			m1++;
		}
		System.out.println(ans);
		
	}
	}