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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int A = std.nextInt();
        int B = std.nextInt();
        int C = std.nextInt();
        int X = std.nextInt();
        int Y = std.nextInt();
        
        int halfCost = C * 2;
        if (halfCost > A + B) {
            long sum = A * X + B * Y;
            System.out.println(sum);
            return;
        }

        long sum = 0;
        if (X < Y) {
            long mixCost = X * C * 2;
            long mixMaxCost = Y * C * 2;
            int restCount = Y - X;
            int restCost = restCount * B;
            sum = Math.min(mixCost + restCost, mixMaxCost);
        } else if (Y < X) {
            long mixCost = Y * C * 2;
            long mixMaxCost = Y * C * 2;
            int restCount = X - Y;
            int restCost = restCount * A;
            sum = Math.min(mixCost + restCost, mixMaxCost);
        }

        System.out.println(sum);
    }
}