Feras bought to his nephew Saleem a new game to help him learning calculating. The game consists of a board with 4 rows and 4 columns with 16 cubes. Every cube has a number from 1 to 16. Let's define the power of a column as the sum of its elements. In the same way, the power of a row is the sum of its elements. Saleem should arrange the cubes in the board such that the power of all columns and all rows are equal. To make the game easier, the nice uncle, Feras, will help him arranging 7 cubes, and Saleem should arrange the rest of the cubes.
Your program will be tested on one or more test cases. The first line of the input will be a single integer T, the number of test cases (1 ≤ T ≤ 100). Then the test cases. Each test case has four lines containing four integers. The j-th number in the i-th line describes the cell (i,j) of the board. If the number is -1 then the cell is empty and you have to fill it, otherwise, uncle Feras has already filled this cell.
For each test case print a line in the following format: "Case c:" where c is the test case number starting from 1 then print the board in four lines every line has four numbers separated by space. If there is more than one solution print the solution that has the smallest order (See the notes below).
in the sample input there is more than one solution:
Solution1:
16 15 2 1
11 6 10 7
4 5 13 12
3 8 9 14
Solution2:
11 6 10 7
16 15 2 1
4 5 13 12
3 8 9 14
but we select solution2 because it has the smallest order when we write the rows in one line.
Solution1: 16 15 2 1 11 6 10 7 4 5 13 12 3 8 9 14
Solution2: 11 6 10 7 16 15 2 1 4 5 13 12 3 8 9 14
## Input
Your program will be tested on one or more test cases. The first line of the input will be a single integer T, the number of test cases (1 ≤ T ≤ 100). Then the test cases. Each test case has four lines containing four integers. The j-th number in the i-th line describes the cell (i,j) of the board. If the number is -1 then the cell is empty and you have to fill it, otherwise, uncle Feras has already filled this cell.
## Output
For each test case print a line in the following format: "Case c:" where c is the test case number starting from 1 then print the board in four lines every line has four numbers separated by space. If there is more than one solution print the solution that has the smallest order (See the notes below).
[samples]
## Note
in the sample input there is more than one solution:Solution1: 16 15 2 1 11 6 10 7 4 5 13 12 3 8 9 14 Solution2:11 6 10 716 15 2 14 5 13 123 8 9 14but we select solution2 because it has the smallest order when we write the rows in one line.Solution1: 16 15 2 1 11 6 10 7 4 5 13 12 3 8 9 14Solution2: 11 6 10 7 16 15 2 1 4 5 13 12 3 8 9 14
**Definitions**
Let $ B \in (\mathbb{Z} \cup \{-1\})^{4 \times 4} $ be the board, where $ B_{i,j} = -1 $ denotes an empty cell, and $ B_{i,j} \in \{1, \dots, 16\} $ denotes a pre-filled number.
Let $ U \subset \{1, \dots, 16\} $ be the set of unused numbers (i.e., numbers not present in $ B $).
Let $ E = \{(i,j) \mid B_{i,j} = -1\} $ be the set of empty cell positions.
**Constraints**
1. $ |E| = 9 $ (7 cells are pre-filled, so 9 are empty).
2. $ |U| = 9 $, and $ U \cap \{B_{i,j} \mid B_{i,j} \neq -1\} = \emptyset $.
3. Each number in $ U $ must be assigned exactly once to a cell in $ E $.
4. Let $ R_i = \sum_{j=1}^4 B_{i,j} $ be the row sum for row $ i $, and $ C_j = \sum_{i=1}^4 B_{i,j} $ be the column sum for column $ j $.
Then:
$$
R_1 = R_2 = R_3 = R_4 = C_1 = C_2 = C_3 = C_4 = S
$$
for some common sum $ S \in \mathbb{Z} $.
5. $ S = \frac{1}{4} \sum_{k=1}^{16} k = 34 $ (since the numbers 1 to 16 sum to 136).
**Objective**
Find a bijection $ f: E \to U $ assigning each empty cell a unique unused number such that all row and column sums equal 34.
Among all valid assignments, select the one that, when the board is read row-wise left-to-right, top-to-bottom, yields the lexicographically smallest 16-tuple.