Description

When John studied the timed automaton, he met the problem about how to trigger the machine. With the problem deeply studied, he found that it can be ascribed to the clock constraints of the timed automaton. The timed automation in question is described below:

The clock variables, or simply clocks, are variables whose values are integers. Of course, time passes at the same rate for all clocks, and any clock can be reset to zero. John uses C to denote the finite set of clocks, and defines the clock constraints for C as follows:

1. All inequalities of the form t # c or c # t are clock constraints, where t is a clock, # is either < or ≤, and c is an integer.
2. If A₁ and A₂ are clock constraints, then A₁ ∧ A₂ is a clock constraint.

John notes that a clock constraint can define several regions in some multidimensional space. He wants to know such regions, so he defines the clock zones recursively as follows.

For simplicity, he let C₀ = C ∪ {x₀}, where x₀ is a reference clock whose value is always 0. The clock zone A can be described by a Difference Bound Matrix D (called a DBM) which is a matrix (D_ij) of size |C₀| × |C₀|. Each D_ij has the form (d_ij, #), where d_ij ∈ Z ∪ {$}, # ∈ {<, ≤}. The value of D_ij can be evaluated in the following form:

For every inequality x_i − x_j # d_ij in clock zone A, let D_ij = (d_ij, #), where x_i and x_j are two clocks. If the bound of x_i − x_j for x_i and x_j is unknown, let D_ij = ($, <).

For example, DBM of the clock zone given by x₁ − x₂ < 2 ∧ 0 < x₂ ≤ 2 ∧ 1 ≤ x₁ is shown below:

The representation of a clock zone by a DBM is not unique. In this example, there are some implied constraints that are not reflected in the DBM. Since x₁ − x₂ < 2 and x₂ ≤ 2, it must be the case x₁ < 4. Since x₀ = 0, the original D₁₀ = ($, <) can be changed into D₁₀ = (4, <). Such adjusting operation is called the tighten operation.

Now John wants to do the similar adjusting operations of difference bounds for all clocks x_i and x_j until further application of this tighten operation does not change the matrix. John obtains the following new canonical difference bound matrix:

Note that some clock zone may contain contrary conditions and has not canonical difference bound matrix.

But John can not obtain a canonical difference bound matrix for a complex clock zone. He asks for your help.

Input

The first line of the input file is a single integer T (1 ≤ T ≤ 20), which is the number of test cases you must process, followed by T test cases:

Each test case consists of several lines. Four integers i, j, d and r are given on each line, representing a constraint x_i − x_j < d or x_i − x_j ≤ d (0 ≤ i, j ≤ m, −10000 < d < 10000). If r = 0, then this line represents an inequality in the form of x_i − x_j < d, otherwise it represents an inequality in the form of x_i − x_j ≤ d. The maximal index m of clocks indicates that the indexes of the clocks are 0, 1, …, m, (1 ≤ m ≤ 100). Note that you have to get the value of m by yourself.

A symbol # given on a single line indicates the end of a test case.

Output

For each test case, first output “Case #:” on a single line, where # is the case number starting from 1. Print a blank line after each test case.

For each test case, output the description of the canonical difference bound matrix. If it doesn’t have a canonical difference bound matrix, print “Canonical DBM does not exist.” (without quotes); If it has a such canonical difference bound matrix, print the matrix in the format as indicated in the sample output. Every element D_ij of the matrix should be written in the form (d_ij, #), where # is either < or <=. If the bound of x_i − x_j for x_i and x_j is unknown, print ($,<) at the position (i, j). Two consecutive elements on each row should be separated by a single space.

1
1 2 2 0
0 2 0 0
2 0 2 1
0 1 -1 1
#

Case 1:
(0,<=) (-1,<=) (0,<)
(4,<) (0,<=) (2,<)
(2,<=) (1,<=) (0,<=)

Source

Shanghai 2006