Problem:
Score : 300 points

Problem Statement

Fennec is fighting with N monsters.

The health of the i-th monster is H_i.

Fennec can do the following two actions:

Attack: Fennec chooses one monster. That monster's health will decrease by 1.

Special Move: Fennec chooses one monster. That monster's health will become 0.

There is no way other than Attack and Special Move to decrease the monsters' health.

Fennec wins when all the monsters' healths become 0 or below.

Find the minimum number of times Fennec needs to do Attack (not counting Special Move) before winning when she can use Special Move at most K times.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 2 \times 10^5

1 \leq H_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
H_1 ... H_N

Output

Print the minimum number of times Fennec needs to do Attack (not counting Special Move) before winning.

Sample Input 1

3 1
4 1 5

Sample Output 1

5

By using Special Move on the third monster, and doing Attack four times on the first monster and once on the second monster, Fennec can win with five Attacks.

Sample Input 2

8 9
7 9 3 2 3 8 4 6

Sample Output 2

0

She can use Special Move on all the monsters.

Sample Input 3

3 0
1000000000 1000000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

Solution in Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

!
! The below module was made in reference to 
! https://ufcpp.net/study/algorithm/col_heap.html
!
module heappq
  
  implicit none
  integer(8),allocatable :: hp(:)

contains 
subroutine heapinit(n)
  
  integer, intent(in) :: n

  allocate( hp(n) ); hp = 0

end subroutine heapinit

subroutine heappush( newnp, newn )
  
  integer, intent(in)    :: newn
  integer(8), intent(in)    :: newnp
  integer :: i,ln
  integer(8) :: tmp
 
  ln = newn
  hp(ln) = newnp
  
  do while ( ln /= 1 )
    i = ln/2
    if ( hp(i) < hp(ln) ) then
      tmp    = hp(ln) 
      hp(ln) = hp(i) 
      hp(i)  = tmp 
    end if
    ln = i
  end do

end subroutine heappush
  
end module heappq

program main
  
  use heappq

  implicit none

  integer    :: n, k, i, j
  integer(8),allocatable :: h(:)
  
  read(*,*) n,k
  allocate( h(n) )
  read(*,*) h

  if ( k >= n ) then 
    print*, "0"
  else if ( k == 0 ) then
    print*, sum(h)
  else

    call heapinit(n) 

    do i = 1, n
      call heappush( h(i), i )
    end do

    h = hp
    print*, h
    
    if( mod( k, 2 ) == 0 ) then
      print*, sum( h(1:k) ) 
    else
      if( h(k) >= h(k+1)) then
        print*, sum( h(1:k) ) 
      else
        print*, sum( h(1:k+1) ) - h(k) 
      end if
    end if

  end if

end program main