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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

    #include <stdio.h>
    #define min(a, b) a < b ? a : b
    int main(void)
    {
        int a, b, ab, x, y;
        scanf("%d%d%d%d%d", &a, &b, &ab, &x, &y);
        if (ab * 2 >= a + b)
        {
            printf("%d\n", a * x + b * y);
            return 0;
        }
        
        int buy_half = min(x, y);
        int left_x = x - buy_half;
        int left_y = y - buy_half;
        
        if (left_x)
        {
            if (ab * 2 < a)
            {
                printf("%d\n", ab * (buy_half + left_x) * 2);
            }
            else
            {
                printf("%d\n", ab * buy_half * 2 + a * left_x);
            }
        }
        if (left_y)
        {
            if (ab * 2 < b)
            {
                printf("%d\n", ab * (buy_half + left_y) * 2);
            }
            else
            {
                printf("%d\n", ab * buy_half * 2 + b * left_y);
            }
        }
        return 0;
    }