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: Wrong Answer)

#include<stdio.h>
int main()
{
int A,B,C,X,Y,z=0,p=0;
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;
}
if(2*C>(X+Y))
{
z=X*A+Y*B;
printf("%d",z);
}

else if((X+Y)>2*C)
{
if(X>Y)
{
z=(X-Y)*A;
p=2*C*Y;
printf("%d",z+p);
}

if(Y>X)
{
z=(Y-X)*B;
p=2*C*X;
printf("%d",z+p);
}

if(X==Y)
{
p=2*C*X;
printf("%d",z+p);
}
}

else if(2*C==(X+Y))
{
z=(X+Y)*C;
printf("%d",z+p);
}

return 0;
}