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(&compare)
    @items = []
    if block_given?
      @compare_proc = compare
    else
      @compare_proc = Proc.new { |a, b| a > b }
    end
  end
  
  def push(x)
    n = @items.size
    @items << x
    while n != 0
      p = (n - 1) / 2
      if @compare_proc.call(@items[n], @items[p])
        @items[p], @items[n] = @items[n], @items[p]
      end
      n = p
    end
  end
  
  def pop
    result = @items[0]
    @items[0] = @items[-1]
    @items.delete_at(-1)

    n = 0
    loop do
      left = 2 * n + 1
      break if left >= @items.size

      right = left + 1
      if (right < @items.size) && @compare_proc.call(@items[right], @items[left])
        c = right
      else
        c = left
      end
      
      break unless @compare_proc.call(@items[c], @items[n])

      @items[n], @items[c] = @items[c], @items[n]
      n = c
    end
    result
  end

  def empty?
    @items.empty?
  end
  
  def size
    @items.size
  end
end

n, m = gets.split.map(&:to_i)
rewards = (m + 1).times.map{Array.new}
n.times do
  a, b = gets.split.map(&:to_i)
  rewards[a] << b if a <= m
end

ans = 0
reward = PriorityQueue.new
1.upto(m) do |i|
  rewards[i].each do |b|
    reward.push(b)
  end
  unless reward.empty?
    ans += reward.pop
  end
end

puts ans