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>

int ans1(int a, int b, int x, int y){
    return a*x + b*y;
}

int ans2(int a, int b, int c, int x, int y){
    if (x >y)
        return 2*y*c + (x - y)*a;
    else
        return 2*x*c + (y - x)*b;
}

int ans3(int c, int x, int y){
    if (x >y)
        return 2*x*c;
    else
        return 2*y*c;
}

int main(void){
    int a, b, c, x, y;
    int anser1, anser2, anser3, true_ans;
    
    /* input */
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    scanf("%d", &x);
    scanf("%d", &y);
    
    anser1 = ans1(a, b, x, y);
    anser2 = ans2(a, b, c, x, y);
    anser3 = ans3(c, x, y);
    
    if (anser1 < anser2 && anser1 < anser3){
        true_ans = anser1;
    }
    else if (anser2 < anser3 && anser2 < anser1){
        true_ans = anser2;
    }
    else {
        true_ans = anser3;
    }
    
    /* output */
    printf("%d \n", true_ans);
        
    return 0;
    }