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>
void MergeSort(long long int x[ ],long long int left,long long int right);
        long long int temp[200001];
int main (void){
int A,B,C;
int X,Y;
scanf("%d %d %d %d %d",&A,&B,&C,&X,&Y);
if(2*C>=A+B){
  printf("%d\n",A*X+B*Y);
}else if(2*C<=A&&2*C<=B){
  if(X>=Y){
    printf("%d\n",C*2*X);
  }else{
    printf("%d\n",C*2*Y);
  }
}else if(2*C<=A||2*C<=B){
  if (2*C<=A) {
    if (X>=Y) {
      printf("%d\n",C*2*X);
    }else{
      printf("%d\n",C*2*X+B*(Y-X));
    }
  }else{
    if (Y>=X) {
      printf("%d\n",C*2*Y);
    }else{
      printf("%d\n",C*2*Y+B*(X-Y));
    }
  }
}else{
  if (X>=Y) {
    printf("%d\n",C*2*Y+A*(X-Y));
  }else{
    printf("%d\n",C*2*X+B*(Y-X));
  }
}
        return 0;
}