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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	//half and half


	public static void main(String[] args){
		int dp1[];//dp[a][b]
		int dp2[];//dp[a][b]
		int A,B,C,X,Y;
		Scanner sc = new Scanner(System.in);
		A= sc.nextInt();
		B= sc.nextInt();
		C= sc.nextInt();
		X= sc.nextInt();
		Y= sc.nextInt();
		dp1=new int[X+1];
		dp2=new int[X+1];
		//if(A>C*2)A=C*2;
		//if(B>C*2)B=C*2;

		//    		dp[0][0]=0;
		//    		dp[1][0]=A;
		//    		dp[0][1]=B;
		//    		dp[1][1]=Math.min(A+B, C*2);


		for(int a=0;a<=X;a++) {
			dp1[a]=A*a;
		}
		for(int b=1;b<=Y;b++) {
			for(int a=0;a<=X;a++) {
				int t1=dp1[a]+B;
				if(a==0) {dp2[a]=t1;continue;}
				int t2=dp2[a-1]+A;
				int t3=dp1[a-1]+C*2;
//				System.err.println("t1\t"+t1);
//				System.err.println("t2\t"+t2);
//				System.err.println("t3\t"+t3);


				dp2[a]=Math.min(t1, Math.min(t2, t3));

			}
			for(int k=0;k<=X;k++) {
				dp1[k]=dp2[k];
				dp2[k]=0;
			}
		}

		System.out.println(dp1[X]);
	}
}