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 Haskell (GHC 8.8.3) (judge verdict: Accepted)

import Data.Maybe
import qualified Data.ByteString.Char8 as BS

readInt = fst . fromJust . BS.readInteger
readIntList = map readInt . BS.words
getIntList = readIntList <$> BS.getLine

data PizzaPrice = PizzaPrice{
priceA :: Integer,
priceB :: Integer,
priceC :: Integer
}

data ReqNum = ReqNum{
quantityX :: Integer,
quantityY :: Integer
}

type Price = Integer

halfAndHalf :: PizzaPrice -> ReqNum -> Price
halfAndHalf pp rn
            | a + b < 2 * c = a * x + b * y
            | x > y = min ((y * 2 * c)  + ( (x - y) * a)) (x * 2 * c)
            | otherwise = min ((x * 2 * c)  + ( (y - x) * b)) (y * 2 * c)
            where
                a = priceA pp
                b = priceB pp
                c = priceC pp        
                x = quantityX rn
                y = quantityY rn                

                
main = do
     [a,b,c,x,y] <- getIntList
     print $ halfAndHalf (PizzaPrice a b c) (ReqNum x y)