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.split.map(&:to_i)

hash = {}
N.times do
  d = gets.split.map(&:to_i)
  next if d[0] > M
  hash[d[0]] = [] unless hash[d[0]]
  hash[d[0]] << d[1]
end

hash.each do |k, v|
  hash[k] = v.sort_by! { |d| -d }[0..M]
end

result = 0
days = 1
bucket = []
while true do
  d = hash[days]
  if d
    bucket += d
    bucket.sort_by! { |d| -d }
  end

  if bucket.size > 0
    result += bucket.shift
  end

  days += 1
  break if days > M
end

p result