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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

#define MIN2(x,y)	((x)<(y)?(x):(y))
#define MIN3(x,y,z)	((x)<(MIN2(y,z)) ? (x):(MIN2(y,z)))

int main(){
  int       a,b,c,x,y;
  long long each, norem1, norem2, rem1, rem2, min1, min2;

  scanf("%d %d %d %d %d", &a, &b, &c, &x, &y);
  
  //1. 各々必要なだけ買う
  each = (a*x) + (b*y);
  
  //2. Aの枚数の方が少ない
  if( x < y ){
    //2-1. 少ない方(A)に合わせてABを買い、足りない方(B)の分を買い足す（余らない）
    rem1 = (2*c*x) + (b*(y-x));
    //2-2. 多い方(B)に合わせてABを買う（余る）
    norem1 = 2*c*y;
    //2-1 と 2-2 を比較
    min1 = MIN2( rem1, norem1 );
  }
  
  //3. Bの枚数の方が少ない
  else if( x > y ){
    //3-1. 少ない方(B)に合わせてABを買い、足りない方(A)の分を買い足す（余らない）
    rem2 = (2*c*y) + (a*(x-y));
    //3-2. 多い方(A)に合わせてABを買う（余る）
    norem2 = 2*c*x;
    //3-1 と 3-2 を比較
    min2 = MIN2( rem2, norem2 );
  }

  //比較
  printf("%lld\n", MIN3( each, min1, min2 ));
  
  return 0;
}