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: Wrong Answer)

def pizza_price_less(a, b, c)
 return true if b*2 < c*2 || a*2 < c*2
 return false
end

def which_buy(x, a, y, b)
  if x < y
    return b
  else
    return a
  end
end

def how_pizza_buy(a, b, c, x, y)
  if a+b < c*2 then
    puts "a+b < c*2"
    return a*x+b*y 
  elsif pizza_price_less(a, b, c)
    puts "elsif pizza_price_less"
    return [x,y].min*c*2+(x-y).abs*which_buy(x, a, y, b)
  else
    puts "else"
    #return min(x,y)*2*c+abs(x-y)*which_buy(x, a, y, b)
    return [x,y].max*c*2
  end
end

a, b, c, x, y = gets.chomp.split.map(&:to_i)
puts how_pizza_buy(a, b, c, x, y)