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 Julia (0.5.0) (judge verdict: Accepted)

parseInt(x) = parse(Int, x)
parseBigInt(x) = parse(BigInt, x)
parseFloat(x) = parse(Float64, x)
parseBigFloat(x) = parse(BigFloat, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
    n,m=parseMap(split(readline()))
    a=sort(parseMap(split(readline())))
    l,r=1,10^10
    x=0                 # 答え
    while r-l>1
        i=l+(r-l)÷2     # 割引券を使う基準となる価格
        d=m             # 割引券の数
        k=0             # 何円必要か
        for j in 1:n
            num=a[j]
            while i<=num
                num÷=2
                d-=1
            end
            k+=num
        end
        if d>=0
            x=k-i÷2*d
            r=i         # d>=0のとき(割引券があまった/ピッタリのとき)は基準を下げる
        else
            l=i         # d<0のとき(割引券が足りなかったとき)は基準を上げる
        end
    end
    println(max(x,0))
end
main()