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 Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

integer N,M,x,y,maxhi
integer,allocatable,dimension(:,:)::A
integer,allocatable,dimension(:)::B
integer(16)ans
read*,N,M
allocate(A(1:M,0:N),B(M))
A=0
B=0
do i=1,N
  read*,x,y
  if(x<=M)then
    B(x)=B(x)+1
    A(x,B(x))=y
  endif
end do
ans=0
do i=1,M
  if(B(i)/=0)then
    call Heapsort(B(i),A(i,1:B(i)))
  end if
  maxhi=0
  do j=1,i
    maxhi=max(maxhi,A(j,B(j)))
  end do
  if(maxhi/=0)then
    do j=1,i
      if(maxhi==A(j,B(j)))then
        B(j)=B(j)-1
        exit
      endif
    end do
  endif
  ans=ans+maxhi
end do

print"(I0)",ans
contains
subroutine heapsort(n,array)
  implicit none
!ここの入力は状況に応じて変更すること
  integer,intent(in) :: n
  integer,intent(inout) :: array(1:n)
  integer::i,k,j,l
  integer:: t

  l=n/2+1
  k=n
  do while(k /= 1)
     if(l > 1)then
        l=l-1
        t=array(L)
     else
        t=array(k)
        array(k)=array(1)
        k=k-1
        if(k == 1) then
           array(1)=t
           exit
        endif
     endif
     i=l
     j=l+l
     do while(j<=k)
        if(j < k)then
           if(array(j) < array(j+1))j=j+1
        endif
        if (t < array(j))then
           array(i)=array(j)
           i=j
           j=j+j
        else
           j=k+1
        endif
     enddo
     array(i)=t
  enddo
  return
end subroutine heapsort
end