Problem:
Score : 300 points

Problem Statement

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.

Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Constraints

1 ≤ A, B, C ≤ 5000

1 ≤ X, Y ≤ 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C X Y

Output

Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.

Sample Input 1

1500 2000 1600 3 2

Sample Output 1

7900

It is optimal to buy four AB-pizzas and rearrange them into two A-pizzas and two B-pizzas, then buy additional one A-pizza.

Sample Input 2

1500 2000 1900 3 2

Sample Output 2

8500

It is optimal to directly buy three A-pizzas and two B-pizzas.

Sample Input 3

1500 2000 500 90000 100000

Sample Output 3

100000000

It is optimal to buy 200000 AB-pizzas and rearrange them into 100000 A-pizzas and 100000 B-pizzas. We will have 10000 more A-pizzas than necessary, but that is fine.

Solution in OCaml (4.02.3) (judge verdict: Accepted)

open Batteries
module EnumL = Enum.Labels
module ListL = List.Labels
module ArrayL = Array.Labels

let dbg0 x = Printf.eprintf "[debug]%s\n" @@ dump x
let dbg1 x = dbg0 x; x

let id = identity

let (++) n m = List.range n `To m
let (++-) n m = if n >= m then List.range n `Downto m else []
let (++^) n m = if n < m then List.range n `To (pred m) else []

let scan fmt = Scanf.sscanf (read_line ()) fmt

let scan_lines n fmt f =
  if n = 0 then []
  else List.map (fun _ -> scan fmt f) (1 ++ n)

let scan_list ?sep cnv =
  let line = read_line () in
  (match sep with
   | None -> List.map String.of_char @@ String.to_list line
   | Some sep -> String.split_on_char sep line)
  |> List.map cnv

let scan_array ?sep cnv = Array.of_list @@ scan_list ?sep cnv

let scan_matrix n m e ?sep conv =
  let arr = Array.make_matrix n m e in
  ListL.iter (0 ++^ n)
    ~f:(fun i -> arr.(i) <- Array.of_list @@ scan_list ?sep conv);
  arr

let rec powerset e =
  match Enum.get e with
  | None -> Enum.singleton @@ Enum.empty ()
  | Some v ->
    let f = powerset e in
    let g = Enum.clone f in
    EnumL.map f ~f:(fun x -> let y = Enum.clone x in push y v; y)
    |>  Enum.append g

let permutations l =
  let rec aux l =
    let rec interleave x = function
      | [] -> [[x]]
      | (hd::tl) as lst ->
        (x::lst) ::
        (ListL.map ~f:(List.cons hd)  @@ interleave x tl)
    in
    match l with
    | [] -> [[]]
    | hd::tl -> List.concat @@ List.map (interleave hd) @@ aux tl in
  let l = List.sort (List.compare Int.compare) @@ aux @@ List.of_enum l in
  List.enum % ListL.map ~f:List.enum @@ l

let intersection l =
  EnumL.filter ~f:(fun x -> EnumL.exists ~f:((=) x) @@ Enum.clone l)

let zip l m =
  let n = min (List.length l) (List.length m) in
  List.combine (List.take n l) (List.take n m)

let lower_bound n f =
  let rec aux l u =
    if u - l > 1 then
      let m = (l + u) / 2 in
      if f m then aux l m
      else aux m u
    else u in aux (-1) n

let between n x m = n <= x && x < m

let (a,b,c,x,y) = scan "%d %d %d %d %d" Tuple5.make

let () =
  [a*x + b*y; 2 * c * (max x y);
   2*c*(min x y) + a*(x- min x y) + b*(y- min x y)]
  |> List.min
  |> Printf.printf "%d\n"