You are given a stream of points on the X-Y plane. Design an algorithm that:

- Adds new points from the stream into a data structure. Duplicate points are allowed and should be treated as different points.
- Given a query point, counts the number of ways to choose three points from the data structure such that the three points and the query point form an axis-aligned square with positive area.

An axis-aligned square is a square whose edges are all the same length and are either parallel or perpendicular to the x-axis and y-axis.

Implement the `DetectSquares` class:

- `DetectSquares()` Initializes the object with an empty data structure.
- `void add(int[] point)` Adds a new point `point = [x, y]` to the data structure.
- `int count(int[] point)` Counts the number of ways to form axis-aligned squares with point `point = [x, y]` as described above.

Example 1:

    Input
    ["DetectSquares", "add", "add", "add", "count", "count", "add", "count"]
    [[], [[3, 10]], [[11, 2]], [[3, 2]], [[11, 10]], [[14, 8]], [[11, 2]], [[11, 10]]]
    Output
    [null, null, null, null, 1, 0, null, 2]

    Explanation
    DetectSquares detectSquares = new DetectSquares();
    detectSquares.add([3, 10]);
    detectSquares.add([11, 2]);
    detectSquares.add([3, 2]);
    detectSquares.count([11, 10]); // return 1. You can choose:
                                   //   - The first, second, and third points
    detectSquares.count([14, 8]);  // return 0. The query point cannot form a square with any points in the data structure.
    detectSquares.add([11, 2]);    // Adding duplicate points is allowed.
    detectSquares.count([11, 10]); // return 2. You can choose:
                                   //   - The first, second, and third points
                                   //   - The first, third, and fourth points

Constraints:

- `point.length == 2`
- `0 <= x, y <= 1000`
- At most `3000` calls in total will be made to `add` and `count`.