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 Ruby (2.3.3) (judge verdict: Accepted)

n,m = gets.split.map(&:to_i)
ab =
  readlines.map do |i|
    i.chomp.split.map(&:to_i)
  end.to_a

days = 1.upto(m).map.to_a
sorted = ab.sort{|a,b| b[1] <=> a[1]}

sum = 0
# 報酬が大きい順に見ていけばGreedyで最適解でそう
# O(mn)を避けるためにbsearchを使うけど、dayで回すと難しいからworkで回す
sorted.each do |a, b|
  break if days.size == 0

  # 報酬もらえるまでの猶予日リストからworkの日数以上の日を消す
  day = days.bsearch_index{|i| i >= a}
  next if day.nil?

  sum += b
  days.delete_at(day)
end

puts sum