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: Time Limit Exceeded)

n,m  = gets.chomp.split.collect{ |item| item.to_i}
list = []
(m+2).times do |i|
  list[i] = [0]
end

n.times do |ni|
  aa,bb = gets.chomp.split.collect{ |item| item.to_i}
  next if aa > m
  list[aa].each_with_index do |bi,idx|
    if bb >= bi
      list[aa].insert(idx,bb)
      break
    end
  end
end

def jointlist(lista,listb,max)
  result = []
  ctime = [max,lista.size+listb.size].min
  ctime.times do |i|
    if lista[0] >= listb[0]
      result.push(lista.shift)
      lista.push(0)
    else
      result.push(listb.shift)
      listb.push(0)
    end
  end
  return result
end
result = 0
canlist = [0]
1.upto(m) do |day|
  canlist = jointlist(canlist,list[day],m-day+1)
  result += canlist.shift
end
puts result