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: Accepted)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in)) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();
            int x = sc.nextInt();
            int y = sc.nextInt();

            // 0. すべてAとBを買うときの金額を初期値とする。
            int ans = a * x + b * y;
            // 1. AとBを2xCに交換し、合計金額の減少or増加を確認する
            // このとき、Cは2枚セットで交換する必要があることに注意
            // 1-1. 合計金額が減少する場合は処理を継続
            // 1-2. 合計金額が減少しない場合は処理を終了
            int max = Math.max(x, y);
            for (int i = 0; i < max; i++) {
                int deltaX = Math.min(x, 1);
                int deltaY = Math.min(y, 1);
                int candidate = ans - (a * deltaX + b * deltaY) + 2 * c * Math.max(deltaX, deltaY);
                if (candidate < ans) {
                    ans = candidate;
                    x -= deltaX;
                    y -= deltaY;
                } else {
                    break;
                }
            }

            System.out.println(ans);
        }
    }

}