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 Heap
  attr_reader :size

  def initialize
    @heap = []
    @size = 0
  end

  def add(n)
    i = @size
    while i > 0 do
      parent_index = (i - 1) / 2
      break if n < @heap[parent_index]

      @heap[i] = @heap[parent_index]
      i = parent_index
    end

    @heap[i] = n
    @size += 1
  end

  def pop
    return if @size <= 0
    min_n = @heap[0]
    @size -= 1
    n = @heap[@size]
    i = 0
    while i * 2 + 1 < @size do
      child_index1 = i * 2 + 1
      child_index2 = i * 2 + 2
      if child_index2 < @size && @heap[child_index2] >= @heap[child_index1]
        child_index1 = child_index2
      end
      break if @heap[child_index1] < n

      @heap[i] = @heap[child_index1]
      i = child_index1
    end
    @heap[i] = n
    min_n
  end

  def min; @heap[0] end
  def values; @heap.first(@size) end
  def inspect; "Heap: #{values}" end
end


N, M = gets.split.map!(&:to_i)
A = 0
B = 1
ab = {}
N.times.map do
  a, b = gets.split.map(&:to_i)
  ab[a] ||= []
  ab[a] << b
end

heap = Heap.new
ans = 0
1.upto(M) do |m|
  ab[m].each { |c| heap.add(c) } if ab[m]
  top = heap.pop()
  ans += top if top
end

puts ans