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>
 
int min(int a,int b){
	if(a<b){
    	return a;
    }else{
    	return b;
    }
}

int max(int a,int b){
	if(a<b){
    	return b;
    }else{
    	return a;
    }
}
 
 
int main(void){
 
  int A=0,B=0,C=0,X=0,Y=0,ans=0;
  scanf("%d%d%d%d%d",&A,&B,&C,&X,&Y);
  
  int pay_a=A*X+B*Y;
  
  int pay_b=0;
  if(X>Y){
  	pay_b=Y*C*2 + A*(X-Y);
  }else{
  	pay_b=X*C*2 + B*(Y-X);
  }
  
  int pay_c=0;
  
  pay_c=C*2*max(X,Y);
  
  int pay=min(pay_a,pay_b);
  pay=min(pay,pay_c);
  
  printf("%d",pay);
  
 

 
 
return 0;
}