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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; ++i)
using ll = long long;
using namespace std;

const int INF = 1e9;
const int MOD = 1e9 + 7;

int main() {

    int A, B, C, X, Y;
    cin >> A >> B >> C >> X >> Y;

    int ans = 0;
    if (A + B >= 2 * C) {
        int mi = min(X, Y);
        ans += mi * 2 * C;
        X -= mi;
        Y -= mi;

        if (!(X == Y && X == 0)) {
            if (X == 0) {
                if (B <= 2 * C)
                    ans += B * Y;
                else
                    ans += 2 * C * Y;
            } else {
                if (A <= 2 * C)
                    ans += A * X;
                else
                    ans += 2 * C * X;
            }
        }
    } else {
        ans += A * X + B * Y;
    }

    cout << ans << endl;

    return 0;
}