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)

fun main(args: Array<String>) = p5()

fun p5() {
    val (a, b, c, x, y) = readLine()!!.split(' ').map { it.toLong() }

    var needA = x
    var needB = y
    var answer = 0L

    if (c * 2 <= a) {
        answer += c * 2 * needA
        needB = Math.max(needB - needA, 0)
        needA = 0
    }

    if (c * 2 <= b) {
        answer += c * 2 * needB
        needA = Math.max(needA - needB, 0)
        needB = 0
    }

    if (c * 2 <= a + b) {
        val ab = Math.min(needA, needB)
        answer += c * 2 * ab
        needA = Math.max(needA - ab, 0)
        needB = Math.max(needB - ab, 0)
    }

    answer += needA * a + needB * b

    println(answer)
}