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>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int max(int a, int b){
  return a >= b ? a : b;
}
int min(int a, int b){
  return b >= a ? a : b;
}
int factorial(int n) {
    if (n > 0) return n*factorial(n - 1);
		else return 1;
}
int compare_up_int(const void *a, const void *b){
   return *(int*)a - *(int*)b;
}
int compare_down_int(const void *a, const void *b){
   return *(int*)b - *(int*)a;
}
int euclid(int a, int b){
  if(a < b){int tmp = a;a = b;b = tmp;}
  int r = a % b;
  if(r < 0) r += b;
  while(r != 0){
    a = b;b = r;r = a % b;
    if(r < 0) r += b;
  }
  return b;
}

int main(){
  int A,B,C,X,Y,count=0,a,b;
  scanf("%d %d %d %d %d",&A,&B,&C,&X,&Y);
  if(A+B > 2*C){
    if(abs(A-B)<C){
      if(X-min(X,Y)<0){
        a=0;
      }
      else a=X-min(X,Y);
      if(Y-min(X,Y)<0){
        b=0;
      }
      else b=Y-min(X,Y);
      printf("%d",(min(X,Y)*C*2 + A*a + B*b));
      return 0;
    }
    else{
      if(X-max(X,Y)<0){
        a=0;
      }
      else a=X-max(X,Y);
      if(Y-max(X,Y)<0){
        b=0;
      }
      else b=Y-max(X,Y);
      printf("%d",(max(X,Y)*C*2 + A*a + B*b));
      return 0;
    }
  }
  printf("%d",A*X + B*Y);
  return 0;
}