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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using E = System.Linq.Enumerable;
 
class Program
 {
        static void Main(string[] args)
        {
          //入力を受け取る
          var ABCXY = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
          int A = ABCXY[0];
          int B = ABCXY[1];
          int C = ABCXY[2];
          int X = ABCXY[3];
          int Y = ABCXY[4];
          
          //枚数が少ない方をXとする
          if(X > Y){
            var tmp1 = X;
            var tmp2 = A;
            X = Y;
            Y = tmp1;
            A = B;
            B = tmp2;
          }
          var Amount = 0L;
          
          //1枚ずつ買った方がABで複製するより得かどうか
          if(A > 2*C && B > 2*C){
            
            //全部Cで作った方が安いという時
            Amount += 2*C*Y;
          }else if(A > 2*C){
            
            //Aの分だけCで作った方がいいとき
            Amount += 2*C*X;
            //残りのBの分を単体で作る
            Amount += B*(Y-X);
          }else if(B > 2*C){
            
            //Bの分をCで複製する(全てCで作る)
            Amount += 2*C*Y;
          }else if(A+B > 2*C){
            
            //X枚だけCで作る
            Amount += 2*C*X;
            
            //Y-X枚のBは単体で作る
            Amount += B*(Y-X);
          }else{
            //全部単体で作る
            Amount += A*X+B*Y;
          }
          Console.WriteLine(Amount);
        }
  
}