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 Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).ok();
    let vec: Vec<&str> = buf.split_whitespace().collect();
    let a: i32 = vec[0].parse().unwrap();
    let b: i32 = vec[1].parse().unwrap();
    let ab: i32 = vec[2].parse().unwrap();
    let mut min_a: i32 = vec[3].parse().unwrap();
    let mut min_b: i32 = vec[4].parse().unwrap();
    let mut ans = 0;

    if min_a > min_b {
        if ab * 2 < a + b{
            while min_b > 0{
                ans = ans + 2 * ab;
                min_b = min_b - 1;
                min_a = min_a - 1;
            }
            if a > 2 * ab{
                while min_a > 0{
                    ans = ans + 2 * ab;
                    min_b = min_b - 1;
                    min_a = min_a - 1;
                }
            }else{
                while min_a > 0 {
                    ans = ans + a;
                    min_a = min_a - 1;
                }
            }
        }else{
            while min_b > 0 {
                ans = ans + b;
                min_b = min_b - 1;
            }
            while min_a > 0 {
                ans = ans + a;
                min_a = min_a - 1;
            }
        }
    }else{
        if ab * 2 < a + b{
            while min_a > 0{
                ans = ans + 2 * ab;
                min_b = min_b - 1;
                min_a = min_a - 1;
            }
            if b > 2 * ab{
                while min_b > 0{
                    ans = ans + 2 * ab;
                    min_b = min_b - 1;
                    min_a = min_a - 1;
                }
            }else{
                while min_b > 0 {
                    ans = ans + b;
                    min_b = min_b - 1;
                }
            }
        }else{
            while min_b > 0 {
                ans = ans + b;
                min_b = min_b - 1;
            }
            while min_a > 0 {
                ans = ans + a;
                min_a = min_a - 1;
            }
        }
    }

    println!("{:?}", ans);

}