Problem:
Score : 400 points

Problem Statement

Takahashi is going to buy N items one by one.

The price of the i-th item he buys is A_i yen (the currency of Japan).

He has M discount tickets, and he can use any number of them when buying an item.

If Y tickets are used when buying an item priced X yen, he can get the item for \frac{X}{2^Y} (rounded down to the nearest integer) yen.

What is the minimum amount of money required to buy all the items?

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_N

Output

Print the minimum amount of money required to buy all the items.

Sample Input 1

3 3
2 13 8

Sample Output 1

9

We can buy all the items for 9 yen, as follows:

Buy the 1-st item for 2 yen without tickets.

Buy the 2-nd item for 3 yen with 2 tickets.

Buy the 3-rd item for 4 yen with 1 ticket.

Sample Input 2

4 4
1 9 3 5

Sample Output 2

6

Sample Input 3

1 100000
1000000000

Sample Output 3

0

We can buy the item priced 1000000000 yen for 0 yen with 100000 tickets.

Sample Input 4

10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 4

9500000000

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

!
! 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, intent(in)    :: newnp
  integer :: i,ln, 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

subroutine heappop( n )
  
  integer :: i,j,ln, tmp
  integer,intent(in) :: n
  
  hp(1) = hp(n)
  
  i = 1 
  do while ( 2*i > n )
    j = 2*i 
    if ( j /= n - 1 .and. hp(j+1) > hp(j) ) j = j + 1
    if ( hp(i) < hp(j) ) then
      tmp    = hp(ln) 
      hp(ln) = hp(i) 
      hp(i)  = tmp 
    end if
    i = j
  end do

end subroutine heappop
 
end module heappq

program main
  
  use heappq  

  implicit none ! 02;01
  integer :: n, m
  integer, allocatable :: a(:)
  integer :: i,j, tmp, tmpo
  integer(8) :: money
  
  read(*,*) n,m
  allocate( a(n) )
  read(*,*) a
  
  call heapinit(n)
  
  do i = 1, n
    call heappush( a(i), i )
  end do

  a = hp 
  do j = 1, m
    tmp = hp(1) / 2 
    call heappop(n)
    call heappush(tmp, n)
  a = hp 
  end do

  a = hp 
  money = 0
  do i = 1, n
    money = money + int( a(i), kind = 8 )
  end do
  print*, money

end program main