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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict";
    
const main = arg => {
    arg = arg.trim().split("\n");
    const A = parseInt(arg[0].split(" ")[0]);
    const B = parseInt(arg[0].split(" ")[1]);
    const C = parseInt(arg[0].split(" ")[2]);
    let X = parseInt(arg[0].split(" ")[3]);
    let Y = parseInt(arg[0].split(" ")[4]);
    
    let answer = 0;
    
    while(X > 0 || Y > 0) {
        if(X !== 0 && Y !== 0) {
            // ABピザを使うべきか
            if(A + B > C * 2) {
                answer += C * 2;
                X--;
                Y--;
            } else {
                if(X > 0) {
                    answer += A;
                    X--;
                } else {
                    answer += B;
                    Y--;
                }
            }
        } else {
            if(X > 0) {
                answer += Math.min(A, C * 2);
                X--;
            } else {
                answer += Math.min(B, C * 2);
                Y--;
            }
        }
        
        // console.log("X:" + X);
        // console.log("Y:" + Y);
    }
    
    console.log(answer);
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));