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.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        long[] price = new long[3];
        for(int i=0;i<3;i++) price[i]=sc.nextLong();
        long[] piece = new long[2];
        piece[0] = sc.nextLong();
        piece[1] = sc.nextLong();
        if(piece[0]>piece[1]){ // piece[0] is always the pizza less needed
            long med = piece[0];
            piece[0] = piece[1];
            piece[1] = med;
            long priceMed = price[0];
            price[0] =price[1];
            price[1] = priceMed;
        }
        long ans = 0;
        ans += Math.min(price[0]*piece[0]+price[1]*piece[0],price[2]*2*piece[0]);
        ans += Math.min(price[1]*(piece[1]-piece[0]),price[2]*2*(piece[1]-piece[0]));
        System.out.print(ans);
    }
}