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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using System.Runtime.CompilerServices;
using static MyUtil;

class MyUtil
{
    public static int[] ReadIntArray()
    {
	return ReadLine().Split().Select(x => int.Parse(x)).ToArray();
    }
}

class Program
{
    public static void Main()
    {
	var tmp = ReadIntArray();
	int n = tmp[0], m = tmp[1];
	int[] d = new int[m+1];

	long ans = 0;
	for (int i = 0; i < n; i++)
	{
	    tmp = ReadIntArray();
	    int a = tmp[0], b = tmp[1];
	    int idx = m - a;
	    for (int j = idx; j >= 0; j--)
	    {
		if (b > d[j])
		{
		    int c = d[j];
		    ans += b - d[j];
		    d[j] = b;
		    b = c;
		}
	    }
	}

	WriteLine(ans);
    }
}