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)

def connect(a,b)
  list = []
  while !a.empty? && !b.empty?
    if a[0] > b[0]
      list.push(a[0])
      a.shift
    else
      list.push(b[0])
      b.shift
    end
  end
  if a.empty?
    list = list + b
  else
    list = list + a
  end
  list
end

n,m = gets.chomp.split(" ").map(&:to_i)

works = Array.new(10**5+1).map{Array.new}
n.times do |i|
  a,b = gets.chomp.split(" ").map(&:to_i)
  works[a].push(b)
end

ans = 0
list = []
for i in 1..m
  works[i].sort!.reverse!
  list = connect(list,works[i])
  if !list.empty?
    ans += list[0]
    list.shift
  end
end
puts ans