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::*;
use std::cmp;

fn read_spc<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn print<T: std::fmt::Display>(x: T) -> () {
    println!("{}", x);
}

fn main() {
    half_and_half();
}

fn half_and_half() {
    let abcxy: Vec<i64> = read_spc();
    let a = abcxy[0];
    let b = abcxy[1];
    let c = abcxy[2];
    let x = abcxy[3];
    let y = abcxy[4];
    let ab = a + b;
    let c2 = c * 2;
    if ab < c2 {
        print(a * x + b * y);
    } else {
        let m = cmp::min(x, y);
        let d = (x - y).abs();
        let mut ans = c2 * m;
        if x == y {
        } else if x < y {
            let m = cmp::min(c2, b);
            ans += m * d;
        } else {
            let m = cmp::min(c2, a);
            ans += m * d;
        }
        print(ans);
    }
}