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)

N, M = gets.split.map(&:to_i)
S = N.times.map { gets.split.map(&:to_i) }.sort_by{|v| -v[1] }
ufs = Array.new(M + 1, &:itself)

class << ufs
  def set_right(l, r)
    loop do
      break if l >= r
      m = self[l]
      self[l] = r
      l = m
    end
    r
  end

  def right(l)
    loop do
      return self.size if l >= self.size
      return l if l == self[l]
      l = self[l]
    end
  end

  def next_clear_index(l)
    r = right(l)
    r >= self.size ? nil : r
  end

  def extend_seq!(l)
    r = next_clear_index(l)
    set_right(l, r + 1) if r
  end
end

puts S.inject(0){|s, (a, b)| ufs.extend_seq!(a) ? s + b : s }