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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int a,b,c,n,m,answer;int s1,s2,s3,s4;
int main()
{  int max(int x,int y);
  int max1(int x,int y,int z);
   while(scanf("%d%d%d%d%d",&a,&b,&c,&n,&m)!=EOF)
   {
      if(n>m)
      {


        s1=(n-(n-m))*2*c+(n-m)*a;
        s2=n*2*c;
      }
        else
        {
            s1=(m-(m-n))*2*c+(m-n)*b;
            s2=m*2*c;
        }
        s3=a*n+b*m;
        answer=max1(s1,s2,s3);
        printf("%d\n",answer);


   }

}
int max(int x,int y)
{
    return x<y?x:y;
}
int max1(int x,int y,int z)
{
    int a;
    a=max(x,y);
    return a<z?a:z;
}