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 <iostream>
#include <string>
#include <vector>
#include <stack>
#include <algorithm>
#include <numeric>
#include <math.h>
using namespace std;

#define N_MAX   (100000)
#define LL_MAX_NUM (1LL<<60)

typedef long long ll;
typedef long long int lli;

int main() {
    int A, B, C ,X, Y, res;
    cin >> A >> B >> C >> X >> Y;
    res = 0;
    if (X >= 1 && Y >= 1) {
        if (A + B > C*2) {
            int minNum = min(X, Y);
            X = X - minNum;
            Y = Y - minNum;
            int res1 = C*minNum*2+A*X+B*Y;
            int res2 = C*minNum*2+C*X*2+C*Y*2;
            res = min(res1, res2); 
        } else {
            res = A*X+B*Y;
        }
    } 
    
    cout << res << endl;
	return 0;
}