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<iostream>
using namespace std;
 
int main(){
int A,B,C,X,Y;
int W,Z;
cin>>A>>B>>C>>X>>Y;
if(X == Y){
  if(A*X+B*Y > C*X*2){
    cout<<C*X*2;
  }
  else{
    cout<<A*X+B*Y;
  }
}

 if(X > Y){
    W = (X+Y)/2;
    W = W*2;
    Z = X - Y;
    if(A*X+B*Y > C*W+A*Z > C*X*2 ){
     cout<<C*X*2;
    }
   else if( A*X+B*Y > C*W+A*Z){
     cout<<C*W+A*Z;
   }
   else{
     cout<<A*X+B*Y;
   }
 }

 if(Y > X){
    W = (X+Y)/2;
    W = W*2;
    Z = Y - X;
    if(A*X+B*Y > C*W+B*Z > C*Y*2){
      cout<<C*Y*2;
    }
    else if(A*X+B*Y > C*W+B*Z){
      cout<<C*W+B*Z;
    }
    else{
      cout<<A*X+B*Y;
    }
 }
    return 0;
}