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 Lua (5.3.2) (judge verdict: Time Limit Exceeded)

--url:https://gist.github.com/LukeMS/89dc587abd786f92d60886f4977b1953
--[[  Priority Queue implemented in lua, based on a binary heap.
Copyright (C) 2017 Lucas de Morais Siqueira <lucas.morais.siqueira@gmail.com>
License: zlib
  This software is provided 'as-is', without any express or implied
  warranty. In no event will the authors be held liable for any damages
  arising from the use of this software.
  Permission is granted to anyone to use this software for any purpose,
  including commercial applications, and to alter it and redistribute it
  freely, subject to the following restrictions:
  1. The origin of this software must not be misrepresented; you must not
     claim that you wrote the original software. If you use this software
     in a product, an acknowledgement in the product documentation would be
     appreciated but is not required.
  2. Altered source versions must be plainly marked as such, and must not be
     misrepresented as being the original software.
  3. This notice may not be removed or altered from any source distribution.
]]--

local floor = math.floor


local PriorityQueue = {}
PriorityQueue.__index = PriorityQueue

setmetatable(
    PriorityQueue,
    {
        __call = function (self)
            setmetatable({}, self)
            self:initialize()
            return self
        end
    }
)


function PriorityQueue:initialize()
    --[[  Initialization.
    Example:
        PriorityQueue = require("priority_queue")
        pq = PriorityQueue()
    ]]--
    self.heap = {}
    self.current_size = 0
end

function PriorityQueue:empty()
    return self.current_size == 0
end

function PriorityQueue:size()
    return self.current_size
end

function PriorityQueue:swim()
    -- Swim up on the tree and fix the order heap property.
    local heap = self.heap
    local floor = floor
    local i = self.current_size

    while floor(i / 2) > 0 do
        local half = floor(i / 2)
        if heap[i][2] < heap[half][2] then
            heap[i], heap[half] = heap[half], heap[i]
        end
        i = half
    end
end

function PriorityQueue:put(v, p)
    --[[ Put an item on the queue.
    Args:
        v: the item to be stored
        p(number): the priority of the item
    ]]--
    --

    self.heap[self.current_size + 1] = {v, p}
    self.current_size = self.current_size + 1
    self:swim()
end

function PriorityQueue:sink()
    -- Sink down on the tree and fix the order heap property.
    local size = self.current_size
    local heap = self.heap
    local i = 1

    while (i * 2) <= size do
        local mc = self:min_child(i)
        if heap[i][2] > heap[mc][2] then
            heap[i], heap[mc] = heap[mc], heap[i]
        end
        i = mc
    end
end

function PriorityQueue:min_child(i)
    if (i * 2) + 1 > self.current_size then
        return i * 2
    else
        if self.heap[i * 2][2] < self.heap[i * 2 + 1][2] then
            return i * 2
        else
            return i * 2 + 1
        end
    end
end

function PriorityQueue:pop()
    -- Remove and return the top priority item
    local heap = self.heap
    local retval = heap[1][1]
    heap[1] = heap[self.current_size]
    heap[self.current_size] = nil
    self.current_size = self.current_size - 1
    self:sink()
    return retval
end

--return PriorityQueue

local n,m=io.read("n","n")
local a=PriorityQueue()
for i=1,n do
    local x=io.read("n")
    a:put(-x,-x)
end
for i=1,m do
    x=a:pop()
    x=-x
    x=x//2
    a:put(-x,-x)
end

local total=0
for i=1,n do
    total=total+a:pop()
end
print(-total)