You have an infinite number of stacks arranged in a row and numbered (left to right) from `0`, each of the stacks has the same maximum capacity.

Implement the `DinnerPlates` class:

- `DinnerPlates(int capacity)` Initializes the object with the maximum capacity of the stacks `capacity`.
- `void push(int val)` Pushes the given integer `val` into the leftmost stack with a size less than `capacity`.
- `int pop()` Returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns `-1` if all the stacks are empty.
- `int popAtStack(int index)` Returns the value at the top of the stack with the given index `index` and removes it from that stack or returns `-1` if the stack with that given index is empty.

Example 1:

    Input
    ["DinnerPlates", "push", "push", "push", "push", "push", "popAtStack", "push", "push", "popAtStack", "popAtStack", "pop", "pop", "pop", "pop", "pop"]
    [[2], [1], [2], [3], [4], [5], [0], [20], [21], [0], [2], [], [], [], [], []]
    Output
    [null, null, null, null, null, null, 2, null, null, 20, 21, 5, 4, 3, 1, -1]

    Explanation:
    DinnerPlates D = DinnerPlates(2);  // Initialize with capacity = 2
    D.push(1);
    D.push(2);
    D.push(3);
    D.push(4);
    D.push(5);         // The stacks are now:  2  4
                                               1  3  5
                                               ﹈ ﹈ ﹈
    D.popAtStack(0);   // Returns 2.  The stacks are now:     4
                                                           1  3  5
                                                           ﹈ ﹈ ﹈
    D.push(20);        // The stacks are now: 20  4
                                               1  3  5
                                               ﹈ ﹈ ﹈
    D.push(21);        // The stacks are now: 20  4 21
                                               1  3  5
                                               ﹈ ﹈ ﹈
    D.popAtStack(0);   // Returns 20.  The stacks are now:     4 21
                                                            1  3  5
                                                            ﹈ ﹈ ﹈
    D.popAtStack(2);   // Returns 21.  The stacks are now:     4
                                                            1  3  5
                                                            ﹈ ﹈ ﹈
    D.pop()            // Returns 5.  The stacks are now:      4
                                                            1  3
                                                            ﹈ ﹈
    D.pop()            // Returns 4.  The stacks are now:   1  3
                                                            ﹈ ﹈
    D.pop()            // Returns 3.  The stacks are now:   1
                                                            ﹈
    D.pop()            // Returns 1.  There are no stacks.
    D.pop()            // Returns -1.  There are still no stacks.

Constraints:

- `1 <= capacity <= 2 * 104`
- `1 <= val <= 2 * 104`
- `0 <= index <= 105`
- At most `2 * 105` calls will be made to `push`, `pop`, and `popAtStack`.