Problem:
Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs.
Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen.
The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

1 \leq Q, H, S, D \leq 10^8

1 \leq N \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90 + 30 + 30 = 150 yen.

Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter.
Thus, you have to buy a 1-liter bottle for 100 yen.

Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10 + 10 + 10 + 10 = 40 yen.

Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942

Solution in JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

main();

function main() {
  var readline = require('readline');
  var reader = readline.createInterface({
      input: process.stdin,
      output: process.stdout
  });
  var lines = [];
  reader.on('line', function (line) {
    lines.push(line);
  });
  reader.on('close', function () {
    var prices   = lines[0].split(/ /);
    var buyLiter = parseInt(lines[1]);
    console.log(caclMinimumMoney(prices, buyLiter));
  });

}

function caclMinimumMoney(prices, buyLiter) {
  var liters = [0.25, 0.5, 1, 2];
  var costs = {};
  for(var i = prices.length - 1; i >= 0; i--) {
    var price = prices[i];
    var liter = liters[i];

    // コスパが良い順に並ぶ
    costs[parseInt(price) / parseFloat(liter)] = {price: price, liter: liter};
  }

  var pay = 0;
  for(var key in costs) {
    var liter = costs[key].liter;
    var price = costs[key].price;

    while (buyLiter >= liter) {
      buyLiter = parseFloat(buyLiter) - parseFloat(liter);
      pay      = parseInt(pay)        + parseInt(price);
    }
  }

  return pay;
}