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 main(){
  int a = 0, b = 0, c = 0, x = 0, y = 0;
  scanf("%d %d %d %d %d", &a, &b, &c, &x, &y);
  int s = 0;
  int z = 0;

  if (x >= y){
    if(a + b <= 2*c){
      s = 1;
    }else if(2*c > a){
      s = 2;
    }else if(a >= 2*c){
      s = 3;
    }
  }else{
    if(a + b <= 2*c){
      s = 1;
    }else if(2*c > b){
      s = 4;
    }else if(b >= 2*c){
      s = 5;
    }
  }

  switch(s){
    case 1:
      z = a*x + b*y;
      break;
    case 2:
      z = a*(x-y) + 2*c*y;
      break;
    case 3:
      z = 2*c*x;
      break;
    case 4:
      z = b*(y-x) + 2*c*x;
      break;
    case 5:
      z = 2*c*y;
      break;
  }
  //printf("case %d ", s);
  printf("%d", z);

  return 0;
}