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 max(int a, int b) {
  if (a > b) return a;
  else return b;
}

int min(int a, int b) {
  if (a > b) return b;
  else return a;
}


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