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 total = 0;
int a, b, c = 0;
int x, y = 0;
int total1, total2, total3 = 0;

void case1(int a,int b,int x,int y){
  total1 = a*x + b*y;
}

void case2(int a,int b,int c,int x,int y){
  if(x==y){
    total2 = a*2;
  }else if(x<y){
    total2 = c*y*2;
  }else if(x>y){
    total2 = c*x*2;
  }
}

void case3(int a,int b,int c,int x,int y){
  if(x==y){
    total3 = a*2;
  }else if(x<y){
    total3 = c*x*2 + (y-x)*b;
  }else if(x>y){
    total3 = c*y*2 + (x-y)*a;
  }
}

void compare(int total1,int total2,int total3){
  total = total1;
  if(total2 < total1) total = total2;
  if(total3 < total) total = total3;
}

int main(void){
  int pizza[4];
  scanf("%d %d %d %d %d",&pizza[0], &pizza[1], &pizza[2], &pizza[3], &pizza[4]);
  
  a = pizza[0];
  b = pizza[1];
  c = pizza[2];
  x = pizza[3];
  y = pizza[4];

  case1(a,b,x,y);
  case2(a,b,c,x,y);
  case3(a,b,c,x,y);

  compare(total1,total2,total3);
  
  printf("%d\n",total);

  return 0;
}