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>

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