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 {

	@SuppressWarnings("rawtypes")
	public static void main(String[] args) {
		@SuppressWarnings("resource")
		Scanner sc = new Scanner(System.in);

		int a = Integer.parseInt(sc.next());
		int b = Integer.parseInt(sc.next());
		int c = Integer.parseInt(sc.next());
		int x = Integer.parseInt(sc.next());
		int y = Integer.parseInt(sc.next());

		/**
		 * 1)A+B<=2Cの場合
		 *  →A*X+B*Yが最低金額
		 * 2)A+B>2Cの場合
		 *  →まず、XとYの小さい方をZとして2C*Z必要
		 *    XとYとの差については、差*A or B or 2C 条件はコード参照
		 *    なお、X=Yなら2C*Xのみ
		 */
		
		int cost = 0;
		// (ai/2)に最も近い数を探索する
		if(a + b <= c * 2) {
			cost += (a * x) + (b * y);
		}else {
			cost += 2 * c * Math.min(x, y);
			if(x > y) {
				if(a <= c * 2) {
					cost += a * (x - y);
				}else {
					cost += 2 * c * (x - y);
				}
			}else if(x < y) {
				if(b <= c * 2) {
					cost += b * (y - x);
				}else {
					cost += 2 * c * (y - x);
				}
			}
		}
		System.out.println(cost);

	}

}