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>
#include <string.h>
#include <stdlib.h>
typedef long long ll;

int main()
{
	int a,b,c,x,y;
	scanf("%d%d%d%d%d",&a,&b,&c,&x,&y);
	int sum = 0;
	if (a + b > c * 2)
	{
		if (x > y)
		{
			sum += c * 2 * y;
			if (a > c*2)
				sum += (x-y) * (c * 2);
			else
				sum += (x-y) * a;
		}
		else if (x < y)
		{
			sum += c * 2 * x;
			if (b > c*2)
				sum += (y - x) * (c * 2);
			else
				sum += (y - x) * b;
		}
		else if (x == y)
			sum =  c * (x + y);
	}
	else
		sum = (a*x)+(b*y);
	printf("%d\n", sum);
	return (0);
}