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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void){
  int A,B,C,X,Y,buff,money;
  char str[100],*p;
  
  fgets(str,sizeof(str),stdin);
  p=strtok(str," \n");
  A=atoi(p);
  p=strtok(NULL," \n");
  B=atoi(p);
  p=strtok(NULL," \n");
  C=atoi(p);
  p=strtok(NULL," \n");
  X=atoi(p);
  p=strtok(NULL," \n");
  Y=atoi(p);
  
  if(X<Y){
    buff=A,A=B,B=buff;
    buff=X,X=Y,Y=buff;
  }
  
  if(2*C<=A+B && 2*C>=A){
    money=2*C*Y;
    money+=A*(X-Y);
  }
  if(2*C<=A+B && 2*C<A){
    money=2*C*X;
  }
  if(2*C>A+B){
    money=A*X+B*Y; 
  }
  
  printf("%d",money);
  
 return 0; 
}