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: Wrong Answer)

#include<stdio.h>

int main(void){
int A, B, C, X, Y;
int nokori;
int money;
   do{
      scanf("%d", &A);
      scanf("%d", &B);
      scanf("%d", &C);
      scanf("%d", &X);
      scanf("%d", &Y);
   }while(A < 1 || A > 5000 || B < 1 || B > 5000 || C < 1 || C > 5000 ||
      X < 1 || X > 100000 || Y < 1 || Y > 100000);


   if(A + B > 2 * C){
   if (X > Y) {
      nokori = X - Y;
      money = nokori*A + 2*C*Y;
   }else if(X < Y){
       nokori = Y - X;
       money = nokori*B + 2*C*X;
   }else{
      nokori = 0;
      money  = 2*C*X;
   }
}else{
   money = A*X + B*Y;
}

   printf("%d", money);





}