"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 sc = new Scanner(System.in);

		//Aの価格
		int A = sc.nextInt();
		//Bの価格
		int B = sc.nextInt();
		//ABの価格
		int C = sc.nextInt();
		//必要なAの枚数
		int X = sc.nextInt();
		//必要なBの枚数
		int Y = sc.nextInt();

		int max = Math.max(X, Y);

		int result = C * 2 * max;

		//Aの枚数
		int x = max;
		//Bの枚数
		int y = max;

		for(int i=0; i<max; i++) {
			//ABを一枚減らしてAかBの片方もしくは両方を一枚増やしたときの合計価格
			int temp = result;
			temp -= 2 * C;
			if(--x<X) {
				x++;
				temp += A;
			}
			if(--y<Y) {
				y++;
				temp += B;
			}

			if(temp >= result) {
				break;
			} else {
				result = temp;
			}
		}

		System.out.println(result);

	}



}
