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>
#include <string.h>
#include <math.h>

#define line_buf 512
#define ll long long

int main(){

    ll int i,j,k;
    ll int A,B,C,X,Y,N,count;
    char line[line_buf];

    fgets(line,sizeof(line),stdin);
    sscanf(line,"%lld %lld %lld %lld %lld",&A,&B,&C,&X,&Y);

    int tmp;

    if(X<Y){    /* X > Yとして進める */
        tmp = X;
        X = Y;
        Y = tmp;
    }

    ll int ans = 0;

    if(A+B < 2*C){      /*ABピザの方が高い*/
        ans = Y*(A+B);
    }

    else{
        ans = Y*2*C;
    }

    /*残りX-Yまい*/

    if(A<2*C){
        printf("%lld",ans+(X-Y)*A);
        return 0;
    }

    else{
        printf("%lld",ans+(X-Y)*2*C);
        return 0;
    }

}