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: Wrong Answer)

#include<stdio.h>
int main()
{
	int A,B,C,X,Y;
	while(1)
	{
		scanf("%d %d %d %d %d",&A,&B,&C,&X,&Y);
		if(A>=1&&B>=1&&C>=1&&A<=5000&&B<=5000&&C<=5000&&X>=1&&Y>=1&&X<=10^5&&Y<=10^5)
		{
			break;
		}
	
	}	
	int x=0,y=0;//当X或Y为0时， 
	
	if(X!=0&&Y!=0)
	{
			if((A+B)>2*C)
	{
		if(X>=Y)
		{
			if(X>=2*C)//买AB
			{
			x=2*(X-Y)*C;		
			}
			else
			{
		    x=(X-Y)*A;
			}
	
		y=Y*2*C;	
		}
		
		else 
		{
			if(Y>=2*C)//买AB
			{
			x=2*(Y-X)*C;		
			}
			else
			{
		    x=(Y-X)*B;
			}
		y=X*2*C;
		}
		printf("%d",x+y);
	}
	else if((A+B)<2*C)
	{
		x=X*A;
		y=Y*B;
		printf("%d",x+y);
	}	
	else
	{
	  if(X>Y)
	  {
	 	x=(X-Y)*A;
	 	y=Y*2*C;
	  } 
	  if(X<Y)
	  {
	 	x=(Y-X)*A;
	 	y=X*2*C;
	  } 
	  if(X==Y)
	  {
	 	x=X*A;
	 	y=Y*B;
	  } 
	printf("%d",x+y);	
    }
	}
	
	else if(X==0||Y==0)
	{
		if(Y==0&&2*C>=A)
		{
			x=X*A; 
			printf("%d",x+y);
		}
		else if(X==0&&2*C>=B)
		{
			y=Y*B;
			printf("%d",x+y);
		}
		else if(Y==0&&2*C<A)
		{
			x=X*2*C;
			printf("%d",x+y);
		}
		else if(Y==0&&2*C<A)
		{
			y=Y*2*C;
			printf("%d",x+y);
		}
		
	}
	
	else if(X==0&&Y==0)
	{
		printf("%d",x+y);
	}

	return 0;
}