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<T: std::str::FromStr>() -> T {
    let stdin = stdin();
    let mut buf = String::new();
	stdin.lock().read_line(&mut buf);
	buf.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().trim().split_whitespace()
        .map(|w| w.parse().ok().unwrap()).collect()
}

fn main() {
    let v = read_vec::<i64>();
    let (a,b,c) = (v[0],v[1],v[2]);
    let (mut x,mut y) = (v[3],v[4]);
    let mut ans = 0;
    if a + b > 2 * c && x > 0 && y > 0 {
        let m = cmp::min(x,y);
        ans += m * 2 * c;
        x -= m;
        y -= m;
    }
    if a >= 2 * c {
        ans += x * 2 * c;
        x = 0;
        y -= x;
    }
    if b >= 2 * c {
        ans += y * 2 * c;
        y = 0;
        x -= y;
    }
    if x > 0 {
        ans += x * a;
    }
    if y > 0 {
        ans += y * b;
    }
    println!("{}",ans);
}