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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
/*
取りうるのは
全部A,Bで揃える
BまでCで揃えてのこりA
A   C             B
全部C
*/
int min(int a,int b){
    if(a<b){
        return a;
    }
    else{
        return b;
    }
}
int max(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);
    int val,val1,val2,val3,val4;
    val1 = X*A + B*Y;
    val2 = Y*2*C;
    if(X>Y){
        val2+=(X-Y)*A;
    }
    val3 = X*2*C;
    if(Y>X){
        val3+=(Y-X)*B;
    }
    val4 = max(X,Y)*2*C;
    val = min(min(val1,val2),min(val3,val4));
    printf("%d\n",val);


    return 0;
}