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: Wrong Answer)


fun main(args:Array<String>) {
    val inpt = readLine()!!.split(" ").map { it.toInt() }
    val prc = Proc(inpt[0], inpt[1], inpt[2], inpt[3], inpt[4])
    println(prc.proc())
}


class Proc(val a:Int, val b:Int, val c:Int, val x:Int, val y:Int) {
    fun proc():Int {
        return getAns(0, Math.max(x,y)*2)
    }

    fun getCost(mixCount:Int):Int {
        var ret = mixCount*c
        ret += Math.max(x - (mixCount/2), 0)*a
        ret += Math.max(y - (mixCount/2), 0)*b
        return ret
    }
    fun getAns(min:Int, max:Int):Int {
        if(max-min <= 2) {
            return listOf(min, (max+min)/2, max).map { getCost(it) }.min()!!
        }
        val idx1 = min + (max-min)/3
        val idx2 = idx1 + (max-min)/3
        val val1 = getCost(min)
        val val2 = getCost(idx1)
        val val3 = getCost(idx2)
        val val4 = getCost(max)
        val minVal = listOf(val1,val2,val3,val4).min()!!
        if(val1 == minVal) {
            return getAns(min, idx1)
        } else if(val2 == minVal) {
            return getAns(min, idx2)
        } else if(val3 == minVal) {
            return getAns(idx1, max)
        } else {
            return getAns(idx2, max)
        }
    }

}