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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		int N = int.Parse(str[0]);
		int M = int.Parse(str[1]);
		int[][]A = new int[N][];
		for(var i=0;i<N;i++){
			string[] str2 = Console.ReadLine().Split();
			A[i] = new int[2]{int.Parse(str2[0]),int.Parse(str2[1])};
		}
		Array.Sort(A,(x,y)=>x[0]-y[0]);
		int na = 0;
		var pq = new PriorityQueue<int,int>();
		long ans = 0;
		for(var i=1;i<=M;i++){
			while(true){
				if(na<N && A[na][0]<=i){
					pq.Add(-A[na][1],na);
					na++;
				} else {
					break;
				}
			}
			if(pq.Count>0){
				var kvpair = pq.Dequeue();
				var p = kvpair.Value;
				ans += A[p][1];
			}
		}
		Console.WriteLine(ans);
	}
}

public class PriorityQueue<TKey, TValue>
{
    SortedDictionary<TKey, Queue<TValue>> dict = new SortedDictionary<TKey, Queue<TValue>>();
    int count = 0;
    public int Count
    {
        get
        {
            return count;
        }
    }
    public void Add(TKey key, TValue value)
    {
        if (!dict.ContainsKey(key))
        {
            dict[key] = new Queue<TValue>();
        }
 
        dict[key].Enqueue(value);
        count++;
    }
    public KeyValuePair<TKey, TValue> Dequeue()
    {
        var queue = dict.First();
        if (queue.Value.Count <= 1)
        {
            dict.Remove(queue.Key);
        }
        count--;
        return new KeyValuePair<TKey, TValue>(queue.Key, queue.Value.Dequeue());
    }
}