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 max(int a, int b);

int main(void){

  int a=0, b=0, c=0;
  int x=0, y=0;
  int sum=0, min=0;
  int i=0, k=0;
  
  scanf("%d %d %d %d %d", &a, &b, &c, &x, &y);

  
  /*if(a+b > 2*c){
    if(x==y){
    price=2*x*c;
      }
    else if(x > y){
    if(2*x > (2*y-(y-x))){
      price=2*y-(y-x);
	}else{
      price=2*x;
	}
      }
    else if(x < y){
    if(2*y > (2*x-(x-y))){
      price=2*y-(x-y);
	}else{
      price=2*y;
	}
      }
    }*/
  
  for(k=0; k<=x; k++){
      sum=2*c*k+max(x-k, 0)*a+max(y-k, 0)*b;
      if(k==0)
	min=sum;
      if(min>sum)
	min=sum;
    }
    printf("%d\n", min);
    
    return 0;
    
}



int max(int a, int b){

  int max=0;
  if(a>=b)
    max=a;
  else
    max=b;

  return max;
}