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.Scanner;

class Main {

  private static long INF = Long.MAX_VALUE / 4;

  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();
    long[] res = new long[3];
    res[0] = Math.min(X, Y) * 2 * C;
    if (X < Y) {
      res[0] += B * (Y - X);
    } else {
      res[0] += A * (X - Y);
    }
    res[1] = A * X + B * Y;
    res[2] = Math.max(X, Y) * 2 * C;
    long ans = INF;
    for (int i = 0; i < res.length; i++) {
      ans = Math.min(res[i], ans);
    }

    System.out.println(ans);
  }

}