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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
/**
*   @Author: MeGaaaaaaaaaa
*   @InitTime: Mon Sep 09 05:03:34 2019
*   @Idea: 
*
*
*
*
* Happy Solving :)
**/
int main(){
#ifndef ONLINE_JUDGE
 //   freopen("in","r",stdin)
 //   freopen("out","w",stdout)
#endif
    int a,b,c,x,y;
    cin>>a>>b>>c>>x>>y;
    int answer=0;
    if(2*c <= b+a){
       int f=min(x,y);
       if(f%2==0){
            answer=f*(2*c);
       }else{
            answer=(f+1)*(2*c);
       }
       x-=f;
       y-=f;
    //   cout<<answer<<endl;
       if(x!=0){
       	  answer+=min((int)((x+(x%2==1))*(2*c)),(int)(x*a));
       }
       if(y!=0){
       	  answer+=min((int)(y+(y%2==1))*(2*c),(int)(y*b));
       }
   }else{
        answer=a*x+y*b;
   }
   cout<<answer<<endl;






}