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 C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
int main(void)
{
	int a, b, c, x, y, price_min, price_ab, price_c_max, price_ab_c;

	scanf("%d %d %d %d %d", &a, &b, &c, &x, &y);
	price_ab = a*x + b*y;
	if(x > y){
		price_c_max = x*2*c;
		price_ab_c  = y*2*c + (x-y)*a;
	}else{
		price_c_max = y*2*c;
		price_ab_c  = x*2*c + (y-x)*b;
	}
	if(price_ab > price_c_max){
		price_min = price_c_max;
		if(price_min > price_ab_c){
			price_min = price_ab_c;
		}		
	}else if(price_ab > price_ab_c){
		price_min = price_ab_c;
	}else{
		price_min = price_ab;
	}
	printf("%d\n", price_min);
	return 0;
}