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: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader buff = null;
		buff = new BufferedReader(new InputStreamReader(System.in));


		int A = 0, B = 0, C = 0;
		double X = 0, Y = 0;
		String[] box = buff.readLine().split(" ");
		A = Integer.parseInt(box[0]);
		B = Integer.parseInt(box[1]);
		C = Integer.parseInt(box[2]);
		X = Integer.parseInt(box[3]);
		Y = Integer.parseInt(box[4]);
		int ans = 0;

		int countAB = 0, countA = 0, countB = 0;
		double pre_X = X, pre_Y = Y;
		while((pre_X > 0) || (pre_Y > 0)){
			if(pre_X >= 0.5 && pre_Y >= 0.5){
				int half = (A/2) + (B/2);
				if(half < C){
					ans += half;
				} else {
					ans += C;
				}
				pre_X -= 0.5;
				pre_Y -= 0.5;
				countAB++;
			} else if(pre_X >= 1){
				ans += A;
				pre_X--;
				countA++;
			} else if(pre_Y >= 1){
				ans += B;
				pre_Y--;
				countB++;
			}
		}

		double over = 0;
		if(X > Y){
			over = X * 2 * C;
		} else {
			over = Y * 2 * C;
		}
		if(ans > over){
			ans = (int)over;
		}

		System.out.println(ans);
	}

	static long f() {
		long ans = 0;


		return ans;
	}
}