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(){
    int A,B,C,X,Y;//A,B,AB's price, A,B wanted
    int PR;//answer
    //int a,b,c;//number of A,B,AB/2
    int damy;
    scanf("%d%d%d%d%d",&A,&B,&C,&X,&Y);//read input
    if(X > Y){//X<Y for all case
        damy = X;
        X = Y;
        Y = damy;
        damy = A;
        A = B;
        B = damy;
    }
    if(2*C - A - B >= 0 && C >= B){
        PR = A * X + B * Y;
    }else if(2*C - A - B >= 0 && C < B){
        PR = A * X + B * Y;
        damy = 2 * C * Y;
        if(damy < PR) PR = damy;
    }else if(2*C - A - B < 0 && C >= B){
        PR = 2 * C * X + B * (Y - X);
    }else{
        PR = 2 * C * Y;
    }
    printf("%d\n",PR);
    return 0;
}