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;
using static System.Console;
using static System.Math;

public class Hello{
    public static void Main(){
        string[] readerstr = ReadLine().Split(' ');
        int[] num = new int[readerstr.Length];
        for(int i=0;i<readerstr.Length;i++){
            num[i] = int.Parse(readerstr[i]);
        }
        long ans = 0;
        //全部AB
        int ab_maxmai = Max(num[3],num[4]);
        int ab = num[2]*2*ab_maxmai;
        ans = Max(ans,ab);
        //ABと残りAかB
        int abandab_kyoutu = Min(num[3],num[4]);
        int abandab = num[2]*2*abandab_kyoutu;
        int abandab_a = (num[3]-abandab_kyoutu)*num[0];
        int abandab_b = (num[4]-abandab_kyoutu)*num[1];
        abandab += abandab_a + abandab_b;
        ans = Min(ans,abandab);
        //全部AかB
        int aorb = num[0]*num[3] + num[1]*num[4];
        ans = Min(ans,aorb);
        
        WriteLine(ans);
    }
}