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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val a = sc.nextInt()
    val b = sc.nextInt()
    val c = sc.nextInt()
    val x = sc.nextInt()
    val y = sc.nextInt()

    var sum = a * x + b * y

    if (c * 2 >= a + b) {
        println(sum)
        return
    }

    var c_count = Math.min(x, y)

    var x_rest = x - c_count
    var y_rest = y - c_count

    c_count *= 2

    while (x_rest != 0 ||  y_rest != 0) {
        if ((Math.min(x_rest, 1) * a + Math.min(y_rest, 1) * b) > c * 2) {
            x_rest = Math.max(0, x_rest - 1)
            y_rest = Math.max(0, y_rest - 1)
            c_count += 2
        } else {
            break
        }
    }

    println(x_rest * a + y_rest * b + c_count * c )

}