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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String[] tokens = in.nextLine().split(" ");
		int A = Integer.parseInt(tokens[0]);
		int B = Integer.parseInt(tokens[1]);
		int C = Integer.parseInt(tokens[2]);
		int X = Integer.parseInt(tokens[3]);
		int Y = Integer.parseInt(tokens[4]);
		// 個別にそろえる方法とABを活用する方法双方を計算して，小さいほうを出力する
		int resultA = X * A + Y * B;
		int resultB = 0;
		if (X > Y) {
			// ABをY個買ってAをX-Y個買う
			resultB = (X - Y) * A + Y * C * 2;
		} else {
			resultB = X * C * 2 + (Y - X) * B;
		}
		int resultC = C * (Math.max(X, Y)) * 2;
		System.out.println(Math.min(Math.min(resultA, resultB), resultC));

		in.close();
	}
}