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)

-- heap
local Heap = {}
function Heap.new(upper_func)
    local self = setmetatable({}, {__index = Heap})
    self.a = {}
    self.upper_func = upper_func
    self.upper = function(k1, k2)
        assert(k1 < k2)
        if k1 == 0 then
            return false
        else
            return upper_func(self.a[k1], self.a[k2])
        end
    end
    self.lower = function(k1, k2)
        return not self.upper(k1, k2)
    end
    return self
end

function Heap:count()
    return #self.a
end

function Heap:_swap(k1, k2)
    local old_k1 = self.a[k1]
    self.a[k1] = self.a[k2]
    self.a[k2] = old_k1
end

function Heap:upheap(newk)
    local ku = newk // 2
    local k = newk
    while ku > 0 and self.lower(ku, k) do
        self:_swap(ku, k)
        k = k//2
        ku = ku//2
    end
end

function Heap:insertq(e)
    table.insert(self.a, e)
    self:upheap(#self.a)
end

function Heap:downheap(newk)
    local k = newk
    while k <= #self.a//2 do
        local kdl = k*2
        local kdr = kdl + 1
        local kd
        if kdr <= #self.a and self.lower(kdl, kdr) then
            kd = kdr
        else
            kd = kdl
        end
        if self.upper(k, kd) then
            break
        end
        self:_swap(k, kd)
        k = kd
    end
end

function Heap:top()
    assert(#self.a >= 1)
    return self.a[1]
end

function Heap:remove_top()
    assert(#self.a >= 1)
    local v = self.a[1]
    self:_swap(1, #self.a)
    table.remove(self.a)
    self:downheap(1)
    return v
end
---
local N, M = io.read("n", "n")
local q = Heap.new(function (x,y) return x > y end)
for i=1,N do
    q:insertq(io.read("n"))
end
for i=1,M do
    local top = q:remove_top()
    top = top / 2
    q:insertq(top)
end
local sum = 0
for i=1,N do
    local top = q:remove_top()
    sum = sum + math.floor(top)
end
print(sum)