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)

#単なる貪欲アルゴリズム
# A+B < 2Cのとき、ABピザを買う価値はない
# A+B > 2Cのとき、ABピザをできるだけ買う方がよい
# A > 2C のとき、Aピザを買う価値はない
# B > 2C のとき、Bピザを買う価値はない
# A+B = 2Cのとき、どちらでもいい
da = gets.split(" ").map{|nn|nn.to_i}
a = da[0]
b = da[1]
c = da[2]
x = da[3]
y = da[4]
ans = 0
if a + b > 2*c
  if x > y
    ans = 2 * c * y
    if a > 2*c
      ans += 2 * c * (x - y)
    else
      ans += a * (x - y)
    end
  else
    ans = 2 * c * x
    if b > 2*c
      ans += 2 * c * (y - x)
    else
      ans += b * (y - x)
    end
  end
else
  ans = a * x + b * y
end
puts ans.to_s