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)

class PriorityQueue
  def initialize(array = [])
    @data = []
    array.each{|a| push(a)}
  end

  def push(element)
    @data.push(element)
    bottom_up
  end

  def pop
    if size == 0
      return nil
    elsif size == 1
      return @data.pop
    else
      min = @data[0]
      @data[0] = @data.pop
      top_down
      return min
    end
  end

  private

  def swap(i, j)
    @data[i], @data[j] = @data[j], @data[i]
  end

  def size
    @data.size
  end

  def parent_idx(target_idx)
    (target_idx - (target_idx.even? ? 2 : 1)) / 2
  end

  def bottom_up
    target_idx = size - 1
    return if target_idx == 0
    parent_idx = parent_idx(target_idx)
    while (@data[parent_idx] > @data[target_idx])
      swap(parent_idx, target_idx)
      target_idx = parent_idx
      break if target_idx == 0
      parent_idx = parent_idx(target_idx)
    end
  end

  def top_down
    target_idx = 0

    # child がある場合
    while (has_child?(target_idx))

      a = left_child_idx(target_idx)
      b = right_child_idx(target_idx)

      # 右側の child が無い場合
      if @data[b].nil?
        if @data[target_idx] > @data[a]
          swap(target_idx, a)
        end
        return
      end

      c = @data[a] < @data[b] ? a : b

      if @data[target_idx] > @data[c]
        swap(target_idx, c)
        target_idx = c
      else
        return
      end
    end
  end

  # @param Integer
  # @return Integer
  def left_child_idx(idx)
    (idx * 2) + 1
  end

  # @param Integer
  # @return Integer
  def right_child_idx(idx)
    (idx * 2) + 2
  end

  # @param Integer
  # @return Boolent
  def has_child?(idx)
    ((idx * 2) + 1) < (@data.size - 1)
  end
end


N, M = gets.chomp.split(' ').map(&:to_i)

AB = []
N.times do |i|
  AB[i] = gets.chomp.split(' ').map(&:to_i)
end

group_by_day = Hash[AB.group_by{|v|v[0]}.sort]

day_fee = Array.new(M){[]}
group_by_day.each do |k,v|
  day_fee[k-1] = v.map{|j| j[1]}
end

# require 'pry'
# binding.pry

ans = 0
pq = PriorityQueue.new
day_fee.each_with_index do |fees, idx|
  break if idx > M - 1
  fees.each do |fee|
    pq.push(fee*(-1))
  end
  ans += pq.pop.to_i
  # p pq
  # p pq.pop*(-1)
end

puts ans*(-1)