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;
public class Main {

	public static void main(String[] args) {
		int a,b,c,x,y = 0;

		Scanner scan = new Scanner(System.in);
		a = Integer.parseInt(scan.next());
		b = Integer.parseInt(scan.next());
		c = Integer.parseInt(scan.next());
		x = Integer.parseInt(scan.next());
		y = Integer.parseInt(scan.next());

/*		for(int i=x;i>=0;i--) {
			int sumx = (a * i) + (x - i)*2*c;
			System.out.println(sumx);
		}
		System.out.println("--------");

		for(int j=y;j>=0;j--) {
			int sumy = (b * j) + (y - j)*2*c;
			System.out.println(sumy);
		}*/

		int num_max = 0;
		if(x > y) {
			num_max = x;
		}else {
			num_max = y;
		}

		int sum = 0;
		for(int i=0;i<=num_max;i++) {

			int sum1 = (x - i)*a;
			int sum2 = (y -i)*b;
			int sum3 = 2*i*c;

			if(sum1 < 0) {
				sum1 = 0;
			}
			if(sum2 < 0) {
				sum2 =0;
			}

			int suma = sum1 + sum2 + sum3;
			if (sum == 0) {
				sum = suma;
			}else if(suma < sum) {
				sum = suma;
			}

		}
		System.out.println(sum);
	}
}