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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void)
{
	int a, b, c, x, y;
	int sum1, sum2, sum3, result;
	int i;

	scanf("%d %d %d %d %d", &a, &b, &c, &x, &y);

	sum1 = (a * x) + (b * y);
	sum2 = c * x * 2;
	if (x < y)
	{
		i = y - x;
		while (0 < i)
		{
			sum2 += b;
			i--;
		}
	}
	sum3 = c * y * 2;
	if (y < x)
	{
		i = x -y;
		while (0 < i)
		{
			sum3 += a;
			i--;
		}
	}

	if (sum1 < sum2)
	{
		result = sum1;

		if (sum3 < sum1)
		{
			result = sum3;
		}
	}
	else
	{
		result = sum2;
		if (sum3 < sum2)
		{
			result = sum3;
		}
	}

	printf("%d", result);

	return 0;
}