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;

namespace abc137_d
{
    static class Program
    {
        static void Main(string[] args)
        {
            var items = Console.ReadLine().ToIntArray();
            // バイトの数
            var n = items[0];
            // 働く日数
            var m = items[1];
            // 支払日 -日後, 給料
            var ab = new SortedDictionary<int, List<int>>();
            for (int i = 0; i < n; i++)
            {
                items = Console.ReadLine().ToIntArray();
                if (!ab.ContainsKey(items[0]))
                {
                    ab.Add(items[0], new List<int>());
                }
                ab[items[0]].Add(items[1]);
            }
            foreach (var item in ab)
            {
                item.Value.Sort();
            }

            // m日後の取得報酬額なので、
            // m日目の仕事→支払日1-0日
            // m-1日目の仕事→支払日2-0日

            // 支払日までの猶予が短い順に
            var money = 0L;
            for (int i = 1; i <= m; i++)
            {
                var a = 0;
                var b = 0;
                foreach (var item in ab)
                {
                    if (item.Key > i)
                    {
                        // 猶予日が過ぎたら一円ももらえない
                        break;
                    }
                    else
                    {
                        // 最大の報酬のみを記録
                        var job = item.Value[item.Value.Count - 1];
                        if (b < job)
                        {
                            a = item.Key;
                            b = job;
                        }
                    }
                }
                if (b > 0)
                {
                    money += b;
                    ab[a].Remove(b);
                    if (ab[a].Count <= 0)
                    {
                        ab.Remove(a);
                    }
                }
            }

            Console.WriteLine($"{money}");
        }

        static int ToInteger(this string s)
        {
            return int.Parse(s);
        }

        static int[] ToIntArray(this string s, char separator = ' ')
        {
            return s.Split(separator).Select(m => m.ToInteger()).ToArray();
        }
    }
}