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.*;
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 sum = 0;
        int max_xy = Max(x,y);
        int min_xy = Min(x,y);
        int max_ab = Max(a,b);
        int max_num, min_num;
        if(max_xy == x){ max_num = a; min_num = b; }
        else           { max_num = b; min_num = a; }

        if( c*2 >= a+b ){
            sum = a*x + b*y;
        }
        if( c*2 < a+b ){
            if( c*2 >= max_num){
                sum = max_num * Math.abs(x-y) + 2 * min_xy * c;
            }
            else{
                sum = 2 * max_xy * c;
            }
        }
        /*if( c <= min){
            sum = c * Max(x,y) * 2;
        }*/
        System.out.println(sum);
    }
    public static int Max (int p, int q){
        int res;
        if( p >= q ) res = p;
        else         res = q;
        return(res);
    }
    public static int Min (int p, int q){
        int res;
        if( p <= q ) res = p;
        else         res = q;
        return(res);
    }
}