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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using System.IO.Pipes;

public class HelloWorld
{
    static bool res = false;
    static long[] avail;
    static long[] memo;
    static public void Main()
    {

        //var N = long.Parse(Console.ReadLine());
        var NM = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
        var A = NM[0];
        var B = NM[1];
        var C = NM[2];
        var X = NM[3];
        var Y = NM[4];


        var MinCost = long.MaxValue;

        for (int c = 0; c <= (long)Math.Max(X,Y)*2; c++)
        {
            var cost = c * C;
            var a = Math.Max(0, X - (c / 2));
            var b = Math.Max(0, Y - (c / 2));
            cost += a * A + b * B;

            MinCost = Math.Min(MinCost, cost);
        }

        Console.WriteLine(MinCost);
    }

    public static long Pow(long a, int n)
    {
        switch (n)
        {
            case 0:
                return 1;
            case 1:
                return a;
            default:
                var p = Pow(a, n / 2);
                return p * p * Pow(a, n % 2);
        }
    }

    public static long PrimeFactorization(long n)
    {
        var res = 1L;

        for (long i = 2; i * i <= n; i++)
        {
            if (n % i != 0) continue;
            long exponents = 0;

            // 割り続ける
            while (n % i == 0)
            {
                exponents++;
                n /= i;
            }

            res *= exponents+1;
        }

        // 素数が残っている場合
        if (n != 1) res *= 2;
        return res;
    }
}