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 OCaml (4.02.3) (judge verdict: Accepted)

module IntMap = Map.Make (struct
  type t = int
  let compare = compare
end)

let add a s =
  IntMap.add a (1 + try IntMap.find a s with Not_found -> 0) s

let remove a s =
  match IntMap.find a s with
  | exception Not_found -> s
  | 1 -> IntMap.remove a s
  | n -> IntMap.add a (n - 1) s

let () = Scanf.scanf "%d %d\n" @@ fun n m ->
  let as_ = Array.init n @@ fun _ -> Scanf.scanf "%d " @@ fun a -> a in
  let rec solve s = function
    | 0 -> s
    | n ->
        let (a, _) = IntMap.max_binding s in
        solve (add (a / 2) (remove a s)) (n - 1) in
  Printf.printf "%d\n" @@
  IntMap.fold (fun a n acc -> acc + a * n)
    (solve (Array.fold_right add as_ IntMap.empty) m) 0