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: Wrong Answer)

def main
  n, m = gets.chomp.split(' ').map(&:to_i)
  jobs = []
  n.times do
    jobs.push(gets.chomp.split(' ').map(&:to_i))
  end
  schedule = (1..m).to_a
  jobs.sort! { |a, b| (b[1] <=> a[1]).nonzero? || b[0] <=> a[0] }
  sum = 0
  jobs.each do |day, reward|
    i = schedule.bsearch_index { |d| d >= day }
    next unless i
    sum += reward
    schedule[i] = -1
  end
  puts sum
end

main if $0 == __FILE__