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>
#include<stdlib.h>
#define  MAXN 10000000
#define  min(a,b) (a)<(b)?(a):(b)

int main(void)
{

	int a, b, c, x, y;
	int sum = 0;
	while (scanf("%d%d%d%d%d", &a, &b, &c, &x, &y) != EOF)
	{
		if (a + b >= 2 * c)
		{
			if (x <= y)
			{
				sum += 2 * x*(c);
				if (b > 2 * c)
					sum += (y - x) * 2 * c;
				else
					sum += (y - x)*b;
			}
			else
			{
				sum += 2 * y*(c);
				if (a > 2 * c)
					sum += (x - y) * 2 * c;
				else
					sum += (x - y)*a;
			}
		}
		else
		{
			if (x <= y)
			{
				sum += x * (a + b);
				if (b > 2 * c)
					sum += (y - x) * 2 * c;
				else
					sum += (y - x)*b;
			}
			else
			{
				sum += y * (a + b);
				if (a > 2 * c)
					sum += (x - y) * 2 * c;
				else
					sum += (x - y)*a;
			}
		}

		printf("%d\n", sum);
		sum = 0;
	}

	return 0;
}