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)

class PQueue
    def initialize(array=[])
        @heap=Array.new
        if !array.empty?
            array.each do |value|
                self.push(value)
            end
        end
    end

    def push(value)
        node=@heap.size
        while node>0
            parent=(node-1)/2
            break if @heap[parent]>=value
            @heap[node]=@heap[parent]
            node=parent
        end
        @heap[node]=value
    end

    def top
        return @heap.first
    end

    def pop
        res=@heap[0]
        value=@heap.pop
        node=0
        while (node*2+1<@heap.size)
            parent=node*2+1
            parent+=1 if parent+1<@heap.size && @heap[parent+1] > @heap[parent]
            break if @heap[node]<=value
            @heap[node]=@heap[parent]
            node=parent
        end
        @heap[node]=value
        return res
    end

    def to_a
        return @heap
    end
end

N,M=gets.split.map(&:to_i)
list=Hash.new()
N.times do |i|
    a,b=gets.split.map(&:to_i)
    list[a] ||= Array.new
    list[a].push(b)
end
que=PQueue.new()
que.push(0)
res=0
1.upto(M) do |i|
    list[i]||=Array.new
    list[i].each do |e|
        que.push(e)
    end
    res+=que.pop
end
puts res