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)

=begin
－－－－－－－－－－－－－－
a, b, c = A, B, AB ピザ値段
X, Y = A, B ピザ必要枚数
dob = A+B と C*2 で安い方の値段を代入
ans = 最小合計
－－－－－－－－－－－－－－
1: dob = A+B と C*2 の安い方(Aピザ1枚 + Bピザ1枚 を作るのに安い組み合わせを探す)
2: X < Y なら
  2.1: ans += dob * X
  2.2: ans += Y - X * (B と C*2 の安い方)
3: X > Y なら
  3.1: ans += dob * Y
  3.2: ans += X - Y * (A と C*2 の安い方)
* X == Y の場合は 何もしない
4: ansを出力
=end

a, b, c, x, y = gets.split.map(&:to_i)
ans = 0
dob = [(a+b), (c*2)].min

if x < y
  min = [b, (c*2)].min
  ans += x * dob
  ans += (y-x) * min
elsif x > y
  min = [a, (c*2)].min
  ans += y * dob
  ans += (x-y) * min
else
  ans += x * dob
end

puts ans