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 Julia (0.5.0) (judge verdict: Accepted)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)
function heap(a::Array{Tuple{Int,Int},1})
	h = [a[1]]
	for i in 2:length(a)
		hpush(h,a[i])
	end
	h
end
function hpush(h::Array{Tuple{Int,Int},1},x::Tuple{Int,Int})
	push!(h,x)
	if length(h) > 1
		l = length(h)
		k = div(l,2)
		while h[k][2]<x[2]&&k>=1
			tmp = h[k]
			h[k] = x
			h[l] = tmp
			k = max(div(k,2),1)
			l = div(l,2)
		end
	end
end

function hpop(h::Array{Tuple{Int,Int},1})
	x = h[1]
	z = pop!(h)
	if !isempty(h)
		h[1] = z
		k = 1
		l = length(h)
		while k <= l
			if 2*k>l
				break
			elseif 2*k+1>l
				if h[2*k][2]<=z[2]
					break
				else
					tmp = h[2*k]
					h[2*k] = z
					h[k] = tmp
					k = 2*k
				end
			else
				if h[2*k][2]<=z[2]&&h[2*k+1][2]<=z[2]
					break
				else
					t = h[2*k][2]>h[2*k+1][2]?2*k:2*k+1
					tmp = h[t]
					h[t] = z
					h[k] = tmp
					k = t
				end
			end
		end
	end
	x
end



function main()
	n,m = readline() |> split |> parseMap
	a = Tuple{Int,Int}[(0,0) for i in 1:n]
	for i in 1:n
		p,q = readline() |> split |> parseMap
		a[i] = (p,q)
	end
	v = zeros(Int,m)
	a = sort(a,by=x->x[1])
	h = Array{Tuple{Int,Int},1}()
	for i in 1:m
		if !isempty(a)
			while a[1][1]<=i
				hpush(h,shift!(a))
				if isempty(a)
					break
				end
			end
		end
		if !isempty(h)
			v[i]=hpop(h)[2]
		end
	end
	println(sum(v))
end

main()