"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.

Examples

Input

1500 2000 1600 3 2


Output

7900


Input

1500 2000 1900 3 2


Output

8500


Input

1500 2000 500 90000 100000


Output

100000000
import java.util.Scanner;
public class Main {

	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		
		int a = stdIn.nextInt(); //aピザの値段
		int b = stdIn.nextInt(); //bピザの値段
		int ab = stdIn.nextInt(); //abピザの値段
		int x = stdIn.nextInt();
		int y = stdIn.nextInt();
		
		int a_b = (ab*2); //aピザ1枚bピザ1枚 
		
		long sum = 0;
		
		//int min = Math.min(a_b, a+b); //abを買うかa,bそれぞれで買うか
		
		int t = 0;
		if(a_b > a+b) {
			t = 1;
		}else {
			t = 2;
		}
		
		
		if(t==2) {
				sum += Math.min(x, y)*a_b;
				if(x > y) {
					sum += (x-y)*a;
				}else {
					sum += (y-x)*b;
				}
			sum = Math.min(sum, a_b*(Math.max(x, y)));
	
		}else {
			sum += (x*a)+(y*b);
		}
		
		System.out.println(sum);
		
		
		
	}

}
