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

public class Main {

	final static long MOD = 1000000007;
	public static void main(String[] args) throws Exception {
		// TODO 自動生成されたメソッド・スタブ

		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String W = in.readLine();
		String[] Ws = W.split(" ");
		long A = Long.parseLong(Ws[0]);
		long B = Long.parseLong(Ws[1]);
		long C = Long.parseLong(Ws[2]);
		long X = Long.parseLong(Ws[3]);
		long Y = Long.parseLong(Ws[4]);

		//Cを利用しない場合の金額
		long ans = A*X+B*Y;

		//Cを利用する枚数を全探索
		int i=0;
		for(;i<=Math.max(X,Y);i++){
			ans=Math.min(ans, Math.max(0,X-i)*A+Math.max(0,Y-i)*B+C*i*2);
		}

		System.out.println(ans);
	}
}