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)

# frozen_string_literal: true

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

# frozen_string_literal: true

# 優先度付きQueue　挿入、削除共に logN 時間で可能！！そのため、logN * M 時間で解ける。
class MaxHeap
  attr_reader :targets

  def initialize(array)
    @targets = {}
    @length = array.length
    @length.times { |i| @targets[i + 1] = array[i] }
  end

  def max_heap(i)
    l = 2 * i
    r = 2 * i + 1

    largest = if l <= @length && @targets[l] > @targets[i]
                l
              else
                i
              end
    largest = r if r <= @length && @targets[r] > @targets[largest]

    return if largest == i

    @targets[i], @targets[largest] = @targets[largest], @targets[i]
    max_heap(largest)
  end

  def build_max_heap
    (@targets.length / 2).downto(1) do |i|
      max_heap(i)
    end
  end

  # 下からつけてどんどん上に
  def insert(val)
    @length += 1
    @targets[@length] = val

    h = @length
    while h > 1 && @targets[h / 2] < @targets[h]
      @targets[h / 2], @targets[h] = @targets[h], @targets[h / 2]
      h /= 2
    end
  end

  def extra_max
    return 0 if @targets[1].nil?

    max = @targets[1]
    @targets[1] = @targets[@length]
    @targets.delete(@length)
    @length -= 1
    max_heap(1)

    max
  end
end

bites = Array.new(M) { [] }
N.times do
  a, b = gets.chomp.split(' ').map(&:to_i)
  next if a > M

  bites[a - 1].push(b)
end

heap = MaxHeap.new([])
cnt = 0
bites.each do |bite|
  bite.each { |b| heap.insert(b) }
  cnt += heap.extra_max
end

puts cnt