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)

# Cがお買い得かどうかをチェック
def good_deal?(a_gap, b_gap)
	return false if a_gap >= 0 && b_gap >= 0
	return true if a_gap < 0 && b_gap < 0

	if (a_gap > 0) && (b_gap.abs - a_gap > 0) || (b_gap > 0) && (a_gap.abs - b_gap > 0)
		true
	else
		false
	end
end

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

a_gap =  (c / 2) - (a / 2)
b_gap =  (c / 2) - (b / 2)

puts (a * x + b * y) unless good_deal?(a_gap, b_gap)

if good_deal?(a_gap, b_gap)
	if (b_gap < 0) && (a_gap < 0)
		puts ([x,y].max * 2 * c)
	elsif (b_gap < 0)
		if (x <= y)
			puts (y*2) * c
		else
			puts (y*2) * c + a * (x-y)
		end
	elsif (a_gap < 0)
		if (y <= x)
			puts (x*2) * c
		else
			puts (x*2) * c + b * (y-x)
		end 
	end
end