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::cmp::max;
use std::cmp::min;

fn main() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    let t :Vec<i32> = s.trim().split_whitespace().map(|x| x.parse::<i32>().unwrap()).collect();

    let (a,b,c,x,y) = (t[0], t[1], t[2], t[3], t[4]);

    let n = 2*max(x,y);
    let mut ans = i32::max_value();
    for i in 0..n+1{
        let mut tmp = i*c;
        let rem_a = max(x-i/2, 0);
        let rem_b = max(y-i/2, 0);
        tmp += a*rem_a + b*rem_b;
        ans = min(ans, tmp);
    }
    println!("{}", ans);
}