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

int minHalf (int a, int b, int c, int x, int y){
    int difference;
    difference = x - y;
    
    int half_cost, both_cost;
    if(difference > 0){
        half_cost = c * (y * 2);
        both_cost = a * difference;
    } else if (difference == 0){
        half_cost = 0;
        both_cost = (a * x) + (b * y);
    } else {
        difference = difference * (-1);
        half_cost = c * (x * 2);
        both_cost = b * difference;
    }
    return half_cost + both_cost;
}

int maxHalf (int c, int x, int y){
    return c * (x > y ? x : y) * 2;
}

typedef struct {
    int both;
    int min_half;
    int max_half;
}ans;

int main(void){
    // Your code here!
    int a,b,c,x,y;
    scanf("%d %d %d %d %d",&a,&b,&c,&x,&y);
    
    ans answer, *p;
    p = &answer;
    
    p->both = Both(a,b,x,y);
    p->min_half = minHalf(a,b,c,x,y);
    p->max_half = maxHalf(c,x,y);
    if(answer.min_half < answer.max_half){
        printf("%d", answer.both < answer.min_half ? answer.both : answer.min_half);
    }else{
        printf("%d", answer.both < answer.max_half ? answer.both : answer.max_half);
    }
    
}