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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#define MIN2(x,y)	((x)<(y)?(x):(y))

int main(){
    int         a,b,c,x,y;
    long long   less, cost;

    scanf("%d %d %d %d %d", &a, &b, &c, &x, &y);

    //AとBで少ない方の枚数
    less = MIN2(x,y);

    //less枚すべてを、2*Cで買った方が安いのか、AとBをそれぞれ買った方が安いのか
    cost = MIN2(2*c*less, (a*less)+(b*less));

    //less枚を超える分を、2*cで買った方が安いのか、単体で買った方が安いのか
    //X > Y
    if     ( x-less > 0 ){ cost += MIN2(2*c*(x-less), a*(x-less)); }
    //X < Y
    else if( y-less > 0 ){ cost += MIN2(2*c*(y-less), b*(y-less)); }

    printf("%lld\n", cost);
    return 0;
}