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
  def initialize(array = [])
    @data = []
    array.each{|a| add(a)}
  end
  
  def add(element)
    @data.push(element)
    bottom_up
  end
  
  # @return Integer
  def peek
    @data[0]
  end

  # @return Integer
  def poll
    if size == 0
      return nil
    elsif size == 1
      return @data.pop
    else
      max = @data[0]
      @data[0] = @data.pop
      top_down
      return max
    end
  end
  
  # @return Integer
  def size
    @data.size
  end

  private

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

  
  def bottom_up
    idx = size - 1
    while(has_parent?(idx))
      parent_idx = parent_idx(idx)
      if @data[parent_idx] < @data[idx]
        swap(parent_idx, idx)
        idx = parent_idx
      else
        return
      end
    end
  end
  
  def top_down
    idx = 0
    while (has_child?(idx))
      a = left_child_idx(idx)
      b = right_child_idx(idx)
      if @data[b].nil?
        c = a
      else
        c = @data[a] >= @data[b] ? a : b
      end
      if @data[idx] < @data[c]
        swap(idx, c)
        idx = c
      else
        return
      end
    end
  end
  

  # @param Integer
  # @return Integer
  def parent_idx(idx)
    (idx - (idx.even? ? 2 : 1)) / 2
  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
  end

  def has_parent?(idx)
    idx > 0
  end
end


n,m=gets.split(' ').map{|i| i.to_i}
matrix = []
n.times do 
  job=gets.split(' ').map{|i| i.to_i}
  if matrix[job[0]].nil?
    matrix[job[0]]=PriorityQueue.new([job[1]])
  else
    matrix[job[0]].add(job[1])
  end
end

total=0
puts total