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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
class Program
{
	static void Main()
	{
	string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
	int a = int.Parse(input[0]);
	int b = int.Parse(input[1]);
    int c = int.Parse(input[2]);
    int x = int.Parse(input[3]);
    int y = int.Parse(input[4]);
      
      long cost = 0;
      long minCost = 0;
    if(x>=y)
    {
       for(int i = 0; i <= x; i++)//Aピザを先に買う
       {
        cost = a*(x-i) + c*2*i + b*(y-i);
         if(cost < minCost || minCost == 0) minCost = cost;
       }
    }
    else
   {
      
      for(int i = 0; i <= y; i++)//Bピザを先に買う
       {
        cost = b*(y-i) + c*2*i + a*(x-i);
         if(cost < minCost || minCost == 0) minCost = cost;
       }
    }

     	 Console.WriteLine(minCost);
	}
}