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 Lua (5.3.2) (judge verdict: Wrong Answer)

A=io.read("n")
B=io.read("n")
C=io.read("n")
X=io.read("n")
Y=io.read("n")
local cost=0
if C*2<=math.min(A,B) then
  cost=C*2*math.max(X,Y)
  elseif C*2>=A+B then
    cost=A*X+B*Y
  elseif C*2<A+B and C*2>=math.max(A,B) then
    if X>Y then
      cost=C*2*math.min(X,Y)+math.abs(X-Y)*A
    else 
      cost=C*2*math.min(X,Y)+math.abs(X-Y)*B
    end
  elseif C*2>math.min(A,B) and C*2<math.max(A,B) then
    if A<B and X<Y or A>B and X>Y then
      cost=C*2*math.max(X,Y)
    elseif A<B and X>Y then
      cost=C*2*Y+A*(X-Y)
    elseif  A>B and X<Y then
      cost=C*2*X+B*(Y-X)
    end
end