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 Ruby (2.3.3) (judge verdict: Accepted)


def solve(a, b, c, x, y)
    res = 10**10 #
     
    0.step([x*2, y*2].max, 2) do |n_c| # C
        n_a = [0, x - n_c/2].max
        n_b = [0, y - n_c/2].max
        s = a*n_a + b*n_b + c*n_c
        # puts "#{s}, #{n_a}, #{n_b}, #{n_c}"
        res = [res, s].min 
    end

    # res = a*x + b*y
    # res = [res, 2*x*c + ((x<y)? (y-x)*b : 0)].min
    # res = [res, 2*y*c + ((x>y)? (x-y)*a : 0)].min
    
    return res 
end

def main()
    a, b, c, x, y = gets.split.map(&:to_i)
    p solve(a, b, c, x, y)
end

main()