"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.*;
import static java.lang.Integer.*;
import static java.lang.Long.*;
//import static java.lang.Math.*;
import static java.lang.System.*;

public class Main {
	public static void main(String[] args) {
		int i,j;
		Scanner sc = new Scanner(in);
		int a = parseInt(sc.next());
		int b = parseInt(sc.next());
		int c = parseInt(sc.next());
		int x = parseInt(sc.next());
		int y = parseInt(sc.next());
		sc.close();
		int mn=(x+y)*Math.max(a, Math.max(b, c));
		int mx=Math.max(x, y);
		int tmp=0; 
		for (i = 0; i <= mx; i++) {
			int tmpx=x-i; 
			int tmpy=y-i;
			tmpx=tmpx<0?0:tmpx;
			tmpy=tmpy<0?0:tmpy;
			tmp=tmpx*a+tmpy*b+i*c*2;
//			out.printf("%d:%d%n", i, tmp);
			if(tmp<mn)mn=tmp;
		}
		out.println(mn);
	}
}