Design an algorithm that accepts a stream of integers and retrieves the product of the last `k` integers of the stream.

Implement the `ProductOfNumbers` class:

- `ProductOfNumbers()` Initializes the object with an empty stream.
- `void add(int num)` Appends the integer `num` to the stream.
- `int getProduct(int k)` Returns the product of the last `k` numbers in the current list. You can assume that always the current list has at least `k` numbers.

The test cases are generated so that, at any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.

Example:

    Input
    ["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"]
    [[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]

    Output
    [null,null,null,null,null,null,20,40,0,null,32]

    Explanation
    ProductOfNumbers productOfNumbers = new ProductOfNumbers();
    productOfNumbers.add(3);        // [3]
    productOfNumbers.add(0);        // [3,0]
    productOfNumbers.add(2);        // [3,0,2]
    productOfNumbers.add(5);        // [3,0,2,5]
    productOfNumbers.add(4);        // [3,0,2,5,4]
    productOfNumbers.getProduct(2); // return 20. The product of the last 2 numbers is 5 * 4 = 20
    productOfNumbers.getProduct(3); // return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40
    productOfNumbers.getProduct(4); // return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0
    productOfNumbers.add(8);        // [3,0,2,5,4,8]
    productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32

Constraints:

- `0 <= num <= 100`
- `1 <= k <= 4 * 104`
- At most `4 * 104` calls will be made to `add` and `getProduct`.
- The product of the stream at any point in time will fit in a 32-bit integer.