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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*; 

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=Math.max(x,y);
        int k=Math.min(x,y);
  		long min=(long)Math.pow(10,18);
        long kei=0;
		for(int e=0;e<=d;e++){
            if(k-e<0){
              if(k==y){
               kei=a*(x-e)+c*2*e;
              }else{
               kei=b*(y-e)+c*2*e;
              }
            }else{
    		kei=a*(x-e)+b*(y-e)+c*2*e;
            }
          min=Math.min(min,kei);
    	}
   

    	System.out.println(min);
    }
}