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.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {

	@SuppressWarnings("resource")
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int a = scanner.nextInt();
		int b = scanner.nextInt();
		int c = scanner.nextInt();
		int x = scanner.nextInt();
		int y = scanner.nextInt();
		Map<Integer, Integer> map = new TreeMap<>();
		int xy[] = new int[2];

		map.put(x, a);
		map.put(y, b);

		int i = 0;
		for (int k : map.keySet()) {
			xy[i] = k;
			i++;
		}

		int ans = 0;
		ans = Math.min(a * xy[0] + b * xy[0], c * xy[0] * 2);

		if (x != y) {
			int t = xy[1] - xy[0];
			ans += Math.min(map.get(xy[1]) * t, c * t * 2);
		}

		System.out.println(ans);

	}
}