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)

# https://atcoder.jp/contests/abc095/tasks/arc096_a



A, B, C, X, Y = gets.split.map(&:to_i)

if C * 2 > A + B
  puts A * X + B * Y
  exit
end

if C <= [A, B].min / 2
  puts C * [X, Y].max * 2
  exit
end

# C を"高い"ピザの必要な枚数を満たすように買う場合
c = A > B ? 2 * X : 2 * Y
cheaper_pizza_count = A > B ? Y : X
ans1 = C * c + [A, B].min * [cheaper_pizza_count - (c / 2), 0].max

# C を"安い"ピザの必要な枚数を満たすように買う場合
c = A > B ? 2 * Y : 2 * X
richer_pizza_count = A > B ? X : Y
ans2 = C * c + [A, B].max * [richer_pizza_count - (c / 2), 0].max

puts [ans1, ans2].min