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 (2.7.1) (judge verdict: Wrong Answer)

n, m = gets.split.map(&:to_i)
ab = n.times.map{|_| gets.split.map(&:to_i)}

g = ab.group_by{|x| x[0]}

ans = 0
c = Array.new
(1..m).each do |i|
  tmp = c
  prefix = []
  g[i].to_a.map{|ab| ab[1]}.sort{|a, b| b <=> a}.each do |i|
    idx = tmp.bsearch_index {|j| j <= i}
    if idx.nil?
      prefix << tmp + [i]
      tmp = []
    else
      tmp_b = tmp
      prefix << tmp.slice(0..idx) + [i]
      tmp = tmp_b.slice(idx..c.length)
    end
  end
  c = prefix + tmp
  c.flatten!
  w = c.shift
  ans += w.to_i
end
p ans