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) {
		// TODO 自動生成されたメソッド・スタブ
		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();
		
		if(a+b<=2*c) {
			System.out.println(a*x+b*y);
			return;
		}
		else if(a>=2*c&&b>=2*c){
			System.out.println(Math.max(x,y)*2*c);
			return;
		}
		else if(a<2*c&&b>=2*c) {
			System.out.println(y*2*c+Math.max(0, x-y)*a);
			return;
		}
		else if(b<2*c&&a>=2*c) {
			System.out.println(x*2*c+Math.max(0, y-x)*b);
			return;
		}
		else if(a<2*c&&b<2*c&&a+b>2*c) {
			int m=Math.min(x, y);
			System.out.println(m*2*c+(y-m)*b+(x-m)*a);
			return;
		}
		
		
	}
}