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: Wrong Answer)

use std::io;
use std::cmp::{max,min};

fn main() {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.trim().split(" ");
    let a = iter.next().unwrap().parse::<i64>().unwrap();
    let b = iter.next().unwrap().parse::<i64>().unwrap();
    let c = iter.next().unwrap().parse::<i64>().unwrap();
    let x = iter.next().unwrap().parse::<i64>().unwrap();
    let y = iter.next().unwrap().parse::<i64>().unwrap();

    if a > 2 * c {
        if b > 2 * c {
            println!("{}", 2 * max(x, y) * c);
        } else {
            println!("{}", 2 * x * c + min(0, y - x) * b);
        }
    } else if b > 2 * c {
        println!("{}", 2 * y * c + min(0, x - y) * a);
    } else if a + b > 2 * c {
        println!("{}", c * 2 * min(x, y) + if x > y {
            a * (x - y)
        } else {
            b * (y - x)
        });
    } else {
        println!("{}", a * x + b * y);
    }
}