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>

using namespace std;

int main()
{
    int a,b,c,y,x,i,j,k;
    scanf("%d%d%d%d%d",&a,&b,&c,&x,&y);

    vector <int> v;
    int sum;

    // only a b
    sum = a*x+b*y;
    v.push_back(sum);

    // only c
    int n= max(x,y);
    sum = n*2*c;
    v.push_back(sum);

    // b direct, a share
    sum = x*2*c;
    i = y - x;
    if(i>0)
    {
        sum = sum + (i*b);
       //printf("i %d ",i);
    }
    v.push_back(sum);

    // A direct, b share
    sum = y*2*c;
    i = x - y;
    if(i>0)
        {
            sum = sum + (i*a);
             //printf("j %d ",i);
        }
        v.push_back(sum);

    sort(v.begin(),v.end());

    printf("%d\n",v[0]);

   return 0;
}