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)

use std::io::Read;
fn main() {
    let mut buf = String::new();

    // 標準入力から全部bufに読み込む
    std::io::stdin().read_to_string(&mut buf).unwrap();

    // 読み込んだStringを空白で分解する
    let mut iter = buf.split_whitespace();

    let a: u64 = iter.next().unwrap().parse().unwrap();
    let b: u64 = iter.next().unwrap().parse().unwrap();
    let c: u64 = iter.next().unwrap().parse().unwrap();

    let x: u64 = iter.next().unwrap().parse().unwrap();
    let y: u64 = iter.next().unwrap().parse().unwrap();

    let mut sum = a * x + b * y;

    let mut xa = x;
    let mut ya = y;
    let mut za = 0;

    loop {
        if xa != 0 {
            xa -= 1
        };
        if ya != 0 {
            ya -= 1
        };

        za += 2;

        let next = xa * a + ya * b + za * c;

        if sum <= next {
            break;
        } else {
            sum = next;
        }
    }

    println!("{}", sum);
}