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)

n,m=gets.split(' ').map{|i| i.to_i}
jobs = []
n.times do 
  jobs<<gets.split(' ').map{|i| i.to_i}
end

jobs.sort!{|a,b| (b[1] <=> a[1]).nonzero? || (b[0] <=> a[0])}
#puts jobs

profit=0
confirmed_jobs=[]
temp_jobs=[]

while jobs.size > 0
  #報酬高い方から一つ選んで追加
  temp_jobs << jobs.shift
  break if temp_jobs.size > m
  #due dateでソート
  temp_jobs.sort{|a,b| b[0] <=> a[0]}
  #制約満たしてるかチェック
  violated=false
  temp_jobs.each_with_index do |job,i|
    if m-i >= job[0]
    else
      violated=true
      break
    end
  end
  #満たしてたら
  if violated
    temp_jobs=confirmed_jobs.dup
  else
    confirmed_jobs=temp_jobs.dup
  end
end
#puts  confirmed_jobs

puts confirmed_jobs.inject(0){|sum, job| sum+=job[1]}