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.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {

	static Scanner sc = new Scanner(System.in);

	public static void main(String[] args){

		solve_abc095_c();
	}

  
  	public static void solve_abc095_c(){

		Scanner sc = new Scanner(System.in);

		int A = sc.nextInt();
		int B = sc.nextInt();
		int C = sc.nextInt();
		int X = sc.nextInt();
		int Y = sc.nextInt();
		
		int ans_a = 0;
		int ans_b = 0;
		int ans_ab = 0;
		int ans_price = 0;

		int tmp_ans_a = 0;
		int tmp_ans_b = 0;
		int tmp_ans_ab = 0;
		int tmp_ans_price = 0;

		// all a & all b
		tmp_ans_a = X;
		tmp_ans_b = Y;
		tmp_ans_ab = 0;
		tmp_ans_price = A * tmp_ans_a + B * tmp_ans_b + C * tmp_ans_ab;
		
		ans_price = tmp_ans_price;
		
		// ab & a
		if(X > Y){
			tmp_ans_a = X -Y;
			tmp_ans_b = 0;
			tmp_ans_ab = Y * 2;
			tmp_ans_price = A * tmp_ans_a + B * tmp_ans_b + C * tmp_ans_ab;
			
			ans_price = Math.min(ans_price, tmp_ans_price);
		}
		// ab & b
		else if(X < Y){
			tmp_ans_a = 0;
			tmp_ans_b = Y - X;
			tmp_ans_ab = X * 2;
			tmp_ans_price = A * tmp_ans_a + B * tmp_ans_b + C * tmp_ans_ab;

			ans_price = Math.min(ans_price, tmp_ans_price);
		}
		
		// all ab
		tmp_ans_a = 0;
		tmp_ans_b = 0;
		tmp_ans_ab = Math.max(X, Y) * 2;
		tmp_ans_price = A * tmp_ans_a + B * tmp_ans_b + C * tmp_ans_ab;

		ans_price = Math.min(ans_price, tmp_ans_price);
		
		System.out.println(ans_price);
		
		sc.close();

	}
}