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 MAX2(x,y)	((x)>(y)?(x):(y))
#define MIN2(x,y)	((x)<(y)?(x):(y))
#define MIN3(x,y,z)	((x)<(MIN2(y,z)) ? (x):(MIN2(y,z)))
#define ABS(x)		((x)<0 ? -(x):(x))

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

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

    //それぞれ買う
    int each = a*x + b*y;

    //A,B少ない方に合わせてCを買って、残りは単体で買う
    int tmp = x>y ? a:b;
    int mix = 2*c*MIN2(x,y) + tmp*ABS(x-y);

    //余ってもいいからCだけで買う
    int only = 2*c*MAX2(x,y);

    //比較
    printf("%d\n", MIN3(each, mix, only));
    return 0;
}