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 Ruby (2.3.3) (judge verdict: Accepted)

class PriorityQueue
  def initialize
    @heap = [nil]
  end

  def up_heap(x)
    @heap << x
    idx = @heap.size - 1

    while idx != 1
      parent = idx / 2
      break if @heap[idx] < @heap[parent]

      @heap[idx], @heap[parent] = @heap[parent], @heap[idx]
      idx = parent
    end
  end

  def down_heap
    size = @heap.size
    return if size == 1
    return @heap.pop if size == 2

    max = @heap[1]
    @heap[1] = @heap.pop
    size -= 1
    idx = 1

    while idx * 2 < size
      l = idx * 2
      r = idx * 2 + 1

      if @heap[r]
        child = @heap[l] > @heap[r] ? l : r
      else
        child = l
      end

      break if @heap[idx] > @heap[child]

      @heap[idx], @heap[child] = @heap[child], @heap[idx]
      idx = child
    end

    max
  end

  def to_a
    @heap.shift
    @heap
  end

  alias_method :<<, :up_heap
  alias_method :push, :up_heap
  alias_method :shift, :down_heap
end

n, m = gets.split.map(&:to_i)
q = PriorityQueue.new
gets.split.map(&:to_i).each { |e| q << e }

m.times { q.push(q.shift / 2) }

puts q.to_a.inject(:+)