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: Time Limit Exceeded)

import java.util.Scanner;
import java.util.stream.Stream;
import java.util.Arrays;
import java.util.StringJoiner;
import java.math.BigInteger;

public class Main {
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);

    int A = sc.nextInt();
    int B = sc.nextInt();
    int C = sc.nextInt();
    int X = sc.nextInt();
    int Y = sc.nextInt();

    // A * X -> AX yen;
    // B * Y -> BY yen;
    // C * Math.max(X, Y) * 2 -> 2 * C * Math.max(X, Y) yen

    // O(n) -> 5000 * 5000 * 5000 -> 125 * 10^9 (NG)
    // O(n) -> 5000 * 5000 -> 25 * 10^6 (OK)

    long minValue = Integer.MAX_VALUE;
    for(int i = 0; i <= X; i++){
      for (int j = 0; j <= Y; j++) {
        long aPizza = A * i;
        long bPizza = B * j;
        // long cPizza = C * Math.max(0, Math.abs(Math.max(X - i, 0) - Math.max(Y - j, 0))) * 2;
        int remainX = X - i > 0 ? X-i : 0;
        int remainY = Y - j > 0 ? Y-j : 0;
        long cPizza = C * Math.max(remainY, remainX) * 2;
        
        minValue = Math.min(minValue, aPizza + bPizza + cPizza);
        // System.out.println("minValue: " + minValue);
      }
    }

    System.out.println(minValue);
 }
}