We'll say a rectangle with coordinates ~(x_1, y_1, z_1, x_2, y_2, z_2)~ is of type X if it is "thin" in the ~x~ dimension, i.e. if ~x_1 = x_2~. Rectangles of types Y and Z are similarly defined.

Define a function ~count(X, Y)~ that does the following:

Select all rectangles of types X and Y and count how many pairs of rectangles there are such that at least one of them is of type X and they share a point.

The overall solution is ~count(X, Y) + count(Y, Z) + count(Z, X)~.

Algorithm for ~count(X, Y)~:

Imagine that the ~y~ coordinate represents time. Let time flow and observe the ~(x, z)~ plane representing space at some moment in time. Rectangles of type X become vertical line segments while rectangles of type Y remain rectangles.

Rectangles of type X have some positive lifetime (because ~y_1 < y_2~), while rectangles of type Y exist at only one instant (because ~y_1 = y_2~).

As time flows, three types of events can occur:

1. A type X rectangle starts
2. A type Y rectangle starts and ends immediately
3. A type X rectangle ends

We sort the events by time and process them one by one.

To count all pairs of rectangles sharing a point, such that one is of type X and another is of type Y, we need to, whenever a rectangle ~(x_1, y, z_1, x_2, y, z_2)~ of type Y comes to life, answer the question "how many vertical line segments in a set intersect the rectangle ~(x_1, z_1, x_2, z_2)~?".

To count all pairs of rectangles sharing a point, such that one is of type X and another is of type X, we need to, whenever a rectangle ~(x_1, y, z_1, x_2, y, z_2)~ of type X comes to life, answer the question "how many vertical line segments in a set intersect the (degenerate) rectangle ~(x, z_1, x, z_2)~?".

Both questions can be quickly answered if we store the starting and ending points of all vertical segments in a 2-dimensional Fenwick tree (a la IOI 2001 mobiles).

Time complexity: ~\mathcal O(N \log N + N \log^2 M)~ where ~M~ is the largest allowed coordinate (~999~ in this problem).