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: Time Limit Exceeded)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int aYen = sc.nextInt();
        int bYen = sc.nextInt();
        int cYen = sc.nextInt();
        int x = sc.nextInt();
        int y = sc.nextInt();

        int cost = Integer.MAX_VALUE;
        for (int i = 0; i <= x; i++) {
            for (int j = 0; j <= y; j++) {
                int abCount = culcCost(i, j, x, y);
                int tmpCost = (aYen*i) + (bYen*j) + (cYen*abCount);

                if (tmpCost < cost) {
                    cost = tmpCost;
                }
            }
        }

        System.out.println(cost);
    }

    public static int culcCost(int aCount, int bCount, int targetACount, int targetBCount) {
        int abCount = 0;

        if (targetACount - aCount > 0) {
            int tmpABCount = (targetACount - aCount)*2;
            if (abCount < tmpABCount) abCount = tmpABCount;
        }

        if (targetBCount - bCount > 0) {
            int tmpABCount = (targetBCount - bCount)*2;
            if (abCount < tmpABCount) abCount = tmpABCount;
        }

        return abCount;
    }
}