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<stdlib.h>
int main(void){
    int A,B,C,X,Y,sum=0,min;
    scanf("%d %d %d %d %d",&A,&B,&C,&X,&Y);
    if(A + B > C * 2){
        if(( A>C && C>B )||( B>C && C>A)){
            if(A > C * 2 || B > C * 2){
                sum = C * 2 * abs(X - Y);
            }else{
                if(X>Y){
                    sum = A * (X - Y);
                }else{
                    sum = B * (Y - X);
                }
            }

            min = X;
            if(min > Y) min = Y;

            sum += min * 2 * C;

        }else{
            if(X > Y){
                sum = X * 2 * C;
            }else{
                sum = Y * 2 * C;
            }
        }
    }else{
        sum = X * A + Y * B;
    }

    printf("%d\n",sum);

    return 0;
}