output_description: Print the minimum amount of money required to prepare X A-pizzas and Y
  B-pizzas.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      A B C X Y
attempt: A, B, C, X, Y = map(int, input().split(" "))
  ans = -1
  for n_ab in range(max(X, Y) + 1):
      n_a = max(X - n_ab, 0)
      n_b = max(Y - n_ab, 0)
      p = n_a * A + n_b * B + n_ab * 2 * C
      ans = p if ans == -1 or p < ans else ans
  print(ans)
  
problem_description: 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.
samples: [{"input": "1500 2000 1600 3 2", "output": "7900\n    \n\nIt is optimal to buy four AB-pizzas and rearrange them into two A-pizzas and\ntwo B-pizzas, then buy additional one A-pizza.\n\n* * *"}, {"input": "1500 2000 1900 3 2", "output": "8500\n    \n\nIt is optimal to directly buy three A-pizzas and two B-pizzas.\n\n* * *"}, {"input": "1500 2000 500 90000 100000", "output": "100000000\n    \n\nIt is optimal to buy 200000 AB-pizzas and rearrange them into 100000 A-pizzas\nand 100000 B-pizzas. We will have 10000 more A-pizzas than necessary, but that\nis fine."}]