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>
#include<stdlib.h>
int max_return(int x,int y){
  if(x>y){
    return x;
  }else{
    return y;
  }
}

int min_return(int x,int y){
  if(x<y){
    return x;
  }else{
    return y;
  }
}

int cost_return(int x,int y,int a,int b){
  if(max_return(x,y)==x){
    return a;
  }
  else{
    return b;
  }
}

int main(){
  int A,B,C,X,Y;
  scanf("%d %d %d %d %d",&A,&B,&C,&X,&Y);
  int a,b;

  if((A+B)>2*C){
    a=C*2*min_return(X,Y)+abs(X-Y)*cost_return(X,Y,A,B);
    b=C*2*max_return(X,Y);
    if(a<b){
      printf("%d",a);
    }
    else{
      printf("%d",b);
    }
  }else{
    printf("%d",X*A+Y*B);
  }
}