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: Wrong Answer)

"use strict";
function Main(input) {
  var parseInputData = parseInput(input);
  resultValue(parseInputData);
}
var parseInput = (input) => {
  console.log(input);
  const lines = input.split(" ");
  //Aピザの価格
  const valuePizzaA = parseInt(lines[0], 10);
  //Bピザの価格
  const valuePizzaB = parseInt(lines[1], 10);
  //ABピザの価格
  const valuePizzaAB = parseInt(lines[2], 10);
  //必要なAピザの枚数
  const needQuantityPizzaA = parseInt(lines[3], 10);
  //必要なBピザの枚数
  const needQuantityPizzaB = parseInt(lines[4], 10);
  return { valuePizzaA: valuePizzaA, valuePizzaB: valuePizzaB, valuePizzaAB: valuePizzaAB, needQuantityPizzaA: needQuantityPizzaA, needQuantityPizzaB: needQuantityPizzaB};
}

//ピザの組み合わせの最安価を出力する
var resultValue = (parseInputData) => {
  parseInputData.valuePizzaAB > (parseInputData.valuePizzaA + parseInputData.valuePizzaB)/2 ?
  console.log(parseInputData.valuePizzaA * parseInputData.needQuantityPizzaA + parseInputData.valuePizzaB * parseInputData.needQuantityPizzaB) : 
  usePizzaABValue(parseInputData)
}

//ピザABを使用する場合の最安価を出力
var usePizzaABValue = (parseInputData) => {
  const needAmountPizzaAB= (parseInputData.needQuantityPizzaA > parseInputData.needQuantityPizzaB ?
    parseInputData.needQuantityPizzaB*2 : parseInputData.needQuantityPizzaA*2);
  console.log(needAmountPizzaAB);
  console.log(parseInputData.needQuantityPizzaA > parseInputData.needQuantityPizzaB ?
    needAmountPizzaAB * parseInputData.valuePizzaAB + parseInputData.valuePizzaA * (parseInputData.needQuantityPizzaA - parseInputData.needQuantityPizzaB) :
    needAmountPizzaAB * parseInputData.valuePizzaAB + parseInputData.valuePizzaB * (parseInputData.needQuantityPizzaB - parseInputData.needQuantityPizzaA)
  )
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));