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 scan = new Scanner(System.in);
		int A = scan.nextInt();
		int B = scan.nextInt();
		int C = scan.nextInt();
		int X = scan.nextInt();
		int Y = scan.nextInt();
		int max = Math.max(X, Y);
		int min = Math.min(X, Y);
		int total = 0;
		int remain = 0;

		if (max == X) { //Aピザの残りを買う必要がある
			remain = A;
		} else {
			remain = B;
		}

		if (2 * A >= 2 * C && 2 * B >= 2 * C) {//Its cheaper to buy AB pizza
			total = total + max * 2 * C;
		} else if (2 * A >= 2 * C && remain == A) {
			total = total + max * 2 * C;
		} else if (2 * B >= 2 * C && remain == B) {
			total = total + max * 2 * C;
		} else if (A + B >= 2 * C) {
			total = total + min * 2 * C + remain * (max - min);
		} else {
			total = A * X + B * Y;
		}
		System.out.println(total);
		scan.close();
	}
}