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 Perl (v5.18.2) (judge verdict: Accepted)

# https://abc095.contest.atcoder.jp/tasks/arc096_a

use strict;

chomp(my $input = <STDIN>);
my ($priceA, $priceB, $priceAB, $numA, $numB) = split(/ /, $input);
my $total = 0;
my $numAB = 0;

# A or Bを優先して買う場合
if ( ($priceA + $priceB) / 2 < $priceAB )
{
    $total = $priceA * $numA + $priceB * $numB;
    print $total . "\n";
    exit 0;
}

# ABを優先して買う場合

# 数量が少ない方を全部満たすだけABを買う
{
    my $min = $numA > $numB ? $numB : $numA;
    $numA -= $min;
    $numB -= $min;
    $numAB = 2 * $min;
}

# 残りについてABのほうがお得ならABで満たす
{
    if ($numA > 0 && $priceA > ($priceAB * 2))
    {
        $numAB += 2 * $numA;
        $numA = 0;
    }

    if ($numB > 0 && $priceB > ($priceAB * 2))
    {
        $numAB += 2 * $numB;
        $numB = 0;
    }
}


$total = $priceA * $numA + $priceB * $numB + $priceAB * $numAB;
print $total . "\n";