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 (1.42.0) (judge verdict: Accepted)

use proconio::{input, fastout};
use proconio::marker::{Chars};
use proconio::marker::Usize1;
use std::cmp::*;

macro_rules! debug {
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

#[fastout]
fn main() {
    input!{
        a: usize,
        b: usize,
        c: usize,
        x: usize,
        y: usize,
    }

    let mut ans: Vec<usize> = Vec::new();
    ans.push(a*x+b*y);
    let more = max(x,y);
    let less = min(x,y);
    for i in 1..more+1 {
        if i<=less {
            ans.push(ans[i-1]-(a+b)+2*c);
        } else {
            if x>=y {
                ans.push(ans[i-1]-a+2*c);
            } else {
                ans.push(ans[i-1]-b+2*c);
            }
        }
        // debug!(ans);
    }

    ans.sort();
    println!("{}", ans[0]);

}