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: Time Limit Exceeded)

program main
  implicit none
  integer :: N, M
  integer(8), allocatable :: A(:)
  integer(8), allocatable :: A2(:)
!  integer(8), allocatable :: A3(:)
  integer :: tmp
  integer :: j2, js
  integer :: i,j
  logical :: fmin
  read(*,*) N,M
  allocate(A(N))
!  allocate(A3(N))
  read(*,*) A(:)
  A2 = iqsort(A) ! sorted
  j2 = N-1
  do i = 1,M
     A2(N) = A2(N) / 2
     !write(*,*) "i:",i
     fmin=.true.
     do j = j2,1,-1
        if (   A2(j) <= A2(N)) then
           ! insert A(N) to A(j+1)
           A(j+1:N) = A2(j+1:N) ! save
           !A2(1:j-2) = A(1:j-2)
           A2(j+1) = A(N)
           A2(j+2:N) = A(j+1:N-1)
           js = j
           fmin = .false.
           exit
        end if
     end do
     if ( fmin ) then
        j = 1
        ! insert A(N) to A(j+1)
        A(j:N) = A2(j:N) ! save
        !A2(1:j-2) = A(1:j-2)
        A2(j) = A(N)
        A2(j+1:N) = A(j:N-1)
        js = 1
     end if
     j2 = js
     !write(*,*) "A2:",A2
  end do
  write(*,'(i0)') sum(A2)
  stop
contains
  recursive function iqsort(ix) result(ires)
    integer(8), allocatable :: ires(:)
    integer(8), intent(in)  :: ix(:)
    integer(8) :: k
    if (size(ix) <= 1) then 
       ires = ix
    else  
       k = ix(size(ix) / 2)
       ires = [iqsort(pack(ix, ix < k)), pack(ix, ix == k), iqsort(pack(ix, ix > k))] 
    end if
  end function iqsort  
end program