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;

fn read_line()-> String {
    let mut s = String::new();
    let _ = io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main() {
	let s = read_line().split_whitespace().map(|e|e.parse().unwrap())
	.collect::<Vec<i32>>();
	let ABC = &s[0..3];
	let X = s[3];
	let Y = s[4];

	let mut c = 0;

	if X > Y {
		c+=((X+Y)+X-Y);
	}else{
		c+=((Y+X)+Y-X);
	}

	if s[0] == s[1] && s[1] == s[2] {
		println!("{}",(X+Y)*s[0]);
	} else if &s[2] == ABC.iter().min().unwrap() {
		println!("{}",c*s[2]);
	}else{
		let m = (X+Y)%2;
		let mut AB = 0;
		if X > Y {
			AB+=(((X+Y)-m)*s[2]);
			AB+=(s[0]*m);
		}else{
			AB+=(((X+Y)-m)*s[2]);
			AB+=(s[1]*m);
		}

		let A = (s[0]*X)+(s[1]*Y);
		if A > AB {
			println!("{}",AB);
		}else{
			println!("{}",A);
		}

	}
}