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(&:to_i)
a = Array.new(n)
b = Array.new(n)
for i in 0..(n-1) 
  a[i], b[i] = gets.split.map(&:to_i)
end

work = Array.new(m).map{Array.new(0)}
for i in 0..(n-1)
  if a[i] <= m
    work[m-a[i]].push(b[i])
  end
end

for i in 0..(m-1)
  if work[i].length > 0
    work[i].sort!
  end
end

sal = 0
for i in 0..(m-1)
  if work[i].length > 0
    sal += work[i].pop
  else
    max = 0
    id = 0
    for j in (i+1)..(m-1)
      if work[j].length > 0
        pops = work[j].pop
        if max < pops
          max = pops
          id = j
        end
        work[j].push(pops)
      end
    end
    sal += max
    if max > 0
      work[id].pop
    end
  end
end
puts sal