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.*;
 
public class Main {
    public static void main(String[] args) {
        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());
        int ans = 0;
        if(a+b <= c*2) {
            ans = a*x + b*y;
        }else {
            if(a*2 <= c*4 && b*2 <= c*4) { 
                if(x >= y) ans = c*2*y + a*(x-y);
                else ans = c*2*x + b*(y-x);
            }else if(a*2 >= c*4 && b*2 <= c*4) {
                if(x >= y) ans = c*2*x;
                else ans = c*2*x + b*(y-x);
            }else if(a*2 <= c*4 && b*2 >= c*4) {
                if(y >= x) ans = c*2*y;
                else ans = c*2*y + b*(x-y);
            }else {
                ans = c*2*max(x,y);
            }
        }
        System.out.println(ans);
    }
    
    static int max(int x, int y) {
        if(x >= y) return x;
        else return y;
    }
    
    static int min(int x, int y) {
        if(x >= y) return y;
        else return x;
    }
}