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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "iostream"
#include "vector"
#include "algorithm"

using namespace std;

typedef long long llint;

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

    cin >> a >> b >> c >> x >> y;

    // int a_half = min(a, c);
    // int b_half = min(b, c):

    // int ans = 0;
    // while(x != 0 && y != 0) {

    //     if (a < c && b < c) {
    //         ans = a * x + b * y;
    //     } if (a > c && b > c) {

    //         if 
    //     }

    // }
    llint min = -1;
    for (int i = 0; i <= x; i++) {
        int ab_num = (x - i) * 2;

        llint price;
        if (y > ab_num / 2) {
            price = i * a + ab_num * c + (y - ab_num / 2) * b;
        } else {
            price = i * a +  ab_num * c;
        }

        if (price < min || min == -1) {
            min = price;
        }
    }

    for (int i = 0; i <= y; i++) {
        int ab_num = (y - i) * 2;

        llint price;
        if (x > ab_num / 2) {
            price = i * b + ab_num * c + (x - ab_num / 2) * c;
        } else {
            price = i * b +  ab_num * c;
        }

        if (price < min || min == -1) {
            min = price;
        }
    }

    cout << min << endl;
}