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: Accepted)

#include <algorithm>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#include <vector>

using namespace std;

#define rep(i, n) for (int64_t i = 0; i < (n); i++)
#define irep(i, n) for (int64_t i = 0; i <= (n); i++)
#define rrep(i, n) for (int64_t i = (n)-1; i >= 0; i--)
#define rirep(i, n) for (int64_t i = n; i >= 0; i--)

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

    x *= 2;
    y *= 2;

    int result = numeric_limits<int>::max();
    irep(k, max(x, y))
    {
        int i = max(0L, (x - k + 1) / 2), j = max(0L, (y - k + 1) / 2);
        int cost = a * i + b * j + c * k;
        result = min(result, cost);
    }
    cout << result << endl;

    return 0;
}