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: Wrong Answer)

class PriorityQueue
  MIN = -(1 << 60)

  attr_reader :size

  def initialize(&block)
    @size = 0
    @heap = Array.new
    @get_key = block || :itself.to_proc
  end

  def empty?
    size == 0
  end

  def pop
    return nil if @size <= 0
    max = @heap[1]
    @heap[1] = @heap[@size]
    @size -= 1
    max_heapify(1)
    max
  end

  def push(x)
    @size += 1
    i = @size
    @heap[i] = x
    while i > 1 && @get_key.call(@heap[i / 2]) < @get_key.call(@heap[i])
      t = @heap[i / 2]
      @heap[i / 2] = @heap[i]
      @heap[i] = t
      i = i / 2
    end
  end

  def top
    return nil if @size <= 0
    @heap[1]
  end

  def to_a
    return [] if @size == 0
    @heap[1..@size]
  end

  private

  def max_heapify(i)
    l = 2 * i
    r = 2 * i + 1
    largest =
      l <= @size && @get_key.call(@heap[l]) > @get_key.call(@heap[i]) ? l : i
    largest = r \
      if r <= @size && @get_key.call(@heap[r]) > @get_key.call(@heap[largest])
    unless largest == i
      t = @heap[i]
      @heap[i] = @heap[largest]
      @heap[largest] = t
      max_heapify(largest)
    end
  end
end


N, M = gets.split.map(&:to_i)
ab_list = N.times.map { gets.split.map(&:to_i) }.select { |(a, b)| a <= M }

pq = PriorityQueue.new do |(a, b)|
  b
end
ab_list.each { |(a, b)| pq.push([a, b]) }
ans = 0
M.times do |i|
  break if pq.empty?
  a, b = pq.pop
  while true
    if i + a <= M
      ans += b
      break
    end
    break if pq.empty?
    a, b = pq.pop
  end
end
puts ans