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 9.2.1) (judge verdict: Accepted)

// template.cpp

// Defines.
//#define DEBUG
#define NUMBER_OF_DIGITS 6

// Includes.
#include <bits/stdc++.h>

#ifdef DEBUG
#include <chrono>
#endif

// Usings.
using namespace std;

// Main function.
int main()
{
#ifdef DEBUG
	chrono::system_clock::time_point start, end;
	
	start = chrono::system_clock::now();
#endif
	
	// settings.
	ios::sync_with_stdio(false);
	
	//cout << fixed;
	//cout << setprecision(NUMBER_OF_DIGITS);
	
	// logic.
	int a, b, c, x, y;
	cin >> a >> b >> c >> x >> y;
	
	int total = 0;
	if ((a + b) >= (2 * c))
	{
		total += (min(x, y) * 2 * c);
		if (x >= y)
		{
			if (a >= (2 * c))
			{
				total += (x - y) * 2 * c;
			}
			else
			{
				total += (x - y) * a;
			}
		}
		else
		{
			if (b >= (2 * c))
			{
				total += (y - x) * 2 * c;
			}
			else
			{
				total += (y -x) * b;
			}
		}
	}
	else
	{
		total += (a * x);
		total += (b * y);
	}
	
	cout << total << endl;

#ifdef DEBUG
	end = chrono::system_clock::now();
	
	double elapsed = chrono::duration_cast<std::chrono::milliseconds>(end-start).count();
	cout << elapsed << "[ms]" << endl;
#endif
	
	return 0;
}