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 Scanf
open Printf

let ($) f x = f x
let scanIntlist n = let list = ref [] in
  for i = 1 to n do
    list := (scanf (if i < n then "%d " else "%d") (fun x -> x))::!list
  done; List.rev !list
    
let string2charlist s = let rec proc i res =
                          try proc (i+1) $ s.[i]::res
                          with Invalid_argument _ -> res
  in List.rev $ proc 0 []
       
let minInt l = let rec proc l res = match l with
      [] -> res
    | x::xs -> if x < res then proc xs x else proc xs res
  in if l == [] then raise $ Failure "null list" else proc l max_int

let maxInt l = let rec proc l res = match l with
      [] -> res
    | x::xs -> if x > res then proc xs x else proc xs res
  in if l == [] then raise $ Failure "null list" else proc l min_int

let rec listSum = function
    [] -> 0
  | x::xs -> x+listSum xs;;

let rec quick_sort = function
    ([] | [_]) as l -> l
  | pivot :: rest ->
     let rec partition left right = function
	 [] -> (quick_sort left) @ (pivot :: quick_sort right)
       | y :: ys -> if pivot < y then partition left (y :: right) ys
                    else partition (y :: left) right ys
     in partition [] [] rest


let calc a b c x y = let p1 = a*x + b*y in
  let p2 = if x < y then c*2*x + b*(y-x) else c*2*y + a*(x-y) in
  let p3 = if x < y then c*2*y else c*2*x in
  minInt [p1;p2;p3];;

scanf "%d %d %d %d %d" $ fun a b c x y -> print_int $ calc a b c x y;;