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) {
		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();
		
		int d = a + b;
		int sum = 0;
		
		if(d <= c*2) {
			sum += Math.min(x,y) * d;
			if(x > y) {
				sum += (x - y) * a;
			}
			else {
				sum += (y - x) * b;
			}
			System.out.println(sum);
		}
		
		if(d > c*2) {
			sum += Math.min(x,y) * 2 * c;
			if(x > y) {
				if(2*c > a) {
				   sum += (x - y) * a;
				}
				if(2*c <= a) {
					sum += (x - y) * 2 * c;
				}
			}
			if(x <= y) {
				if(2*c> b) {
					   sum += (y - x) * b;
					}
			    if(2*c <= b){
						sum += (y - x) * 2 * c;
					}
				}
			System.out.println(sum);
		}
		

	}

}