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: Accepted)

using System;

class HalfAndHalf{
    static void Main(string[] args){
        var input = Console.ReadLine().Split(' ');
        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]);


        if(a * 2 + b * 2 < c * 4){
            // これはそれぞれ一枚ずつ買ったほうがいい
            Console.WriteLine("{0}", a*x + b*y);
            return;
        }

        var sum = 0;
        var nokoriA = x;
        var nokoriB = y;
        // まずCを2枚でAB1枚ずつつくる
        if(x<y){
            sum = sum + c*2*x;
            nokoriA = nokoriA - x;
            nokoriB = nokoriB - x;
            if(b*1 > c*2){
                // Bを1枚買うよりCを2枚買ったほうがいい
                sum = sum + 2*c*nokoriB;
            }else{
                // のこりのBをBで埋める
                sum = sum + b*nokoriB;
            }
        }else{
            sum = sum + c*2*y;
            nokoriA = nokoriA - y;
            nokoriB = nokoriB - y;
            
            if(a*1 > c*2){
                // Aを1枚買うよりCを2枚買ったほうがいい
                sum = sum + 2*c*nokoriA;
            }else{
                // のこりのAをAで埋める
                sum = sum + a*nokoriA;
            }
        }


        Console.WriteLine(sum);
    }
}