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.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        final int A = Integer.parseInt(sc.next());
        final int B = Integer.parseInt(sc.next());
        final int C = Integer.parseInt(sc.next());
        int x = Integer.parseInt(sc.next());
        int y = Integer.parseInt(sc.next());
        long ans = 1000000000;
        int min = Math.min(x, y);
        int max = Math.max(x, y);
        long[] sum = new long[3];
        
        sum[0] = A*x + B*y;
        sum[1] = 2 * C * max;
        if (x >= y) {
            sum[2] = 2 * C * min + (max - min) * A;
        } else {
            sum[2] = 2 * C * min + (max - min) * B;
        }
        for (int i = 0; i < 3; i++) {
            ans = Math.min(ans, sum[i]);
        }
        
        out.println(ans);
        
        out.flush();
    }
}