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>
#include <math.h>
int main(void){
  int A,B,C,X,Y;
  scanf("%d%d%d%d%d",&A,&B,&C,&X,&Y);
  int min = X,max = Y;
  int judge = 0;
  if(X > Y){
    max = X;
    min = Y;
    judge += 10;
  }
  int min_v = A,max_v = B;
  if(A > B){
    max_v = A;
    min_v = B;
    judge += 1;
  }

  if(A + B < 2 * C){
    printf("%d", A * X + B * Y);
  }else if(min_v >= 2 * C){
    printf("%d",C * max);
  }else{
    int ans = 0;
    ans += C * 2 *min;
    switch(judge){
      case 00:  if(2 * C > B){
                  ans += (max - min) * B;
                }else{
                  ans += (max - min) * 2 * C;
                }
                break;
      case 01:  ans += (max - min) * A; break;
      case 10:  ans += (max - min) * B; break;
      case 11:  if(2 * C > A){
                  ans += (max - min) * A;
                }else{
                  ans += (max - min) * 2 * C;
                }
                break;
    }
    printf("%d",ans);
  }
  return 0;
}