"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.
A,B,C,X,Y = map(int,input().split())
if A<=C  and B<=C:
  print(A*X+B*Y)
  
elif C<=A and C<=B:
  if X<=Y:
    a = 2*C*X+(Y-X)*B
    b = 2*C*Y 
    print(min(a,b))
  else:
    a = 2*C*Y+(X-Y)*A
    b = 2*C*X
    print(min(a,b))
    
elif A<=C and B>C:
  print(A*X+min(B,2*C)*Y)
  
else:
  print(B*Y + min(A,2*C)*X)
A,B,C,X,Y = map(int,input().split())
if A<C  and B<C:
  print(A*X+B*Y)
  
elif C<=A and C<=B:
  if X<=Y:
    a = 2*C*X+(Y-X)*B
    b = 2*C*Y 
    print(min(a,b))
  else:
    a = 2*C*Y+(X-Y)*A
    b = 2*C*X
    print(min(a,b))
    
elif A<=C and B>=C:
  if X<=Y:
    a = 2*C*X+(Y-X)*B
    b = A*X +B*Y
    print(min(a,b))
  else:
    a = 2*C*Y+(X-Y)*A
    b = A*X +B*Y
    print(min(a,b))
else:
  if X<=Y:
    a = 2*C*X+(Y-X)*B
    b = A*X +B*Y
    print(min(a,b))
  else:
    a = 2*C*Y+(X-Y)*A
    b = A*X +B*Y
    print(min(a,b))