Given a matrix (a) of size (n \times m), each cell of which contains a non-negative integer. The integer lying at the intersection of the (i)-th row and the (j)-th column of the matrix is called (a_{i,j}). Let's define (f(i)) and (g(j)) as the XOR of all integers in the (i)-th row and the (j)-th column, respectively. In one operation, you can either: Select any row (i), then assign (a_{i,j} := g(j)) for each (1 \le j \le m); or Select any column (j), then assign (a_{i,j} := f(i)) for each (1 \le i \le n). In this example, as we apply an operation on column (2), all elements in this column are changed: (a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0) (a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3) (a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1) (a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29) You can apply the operations any number of times. Then, we calculate the (\textit{beauty}) of the final matrix by summing the absolute differences between all pairs of its adjacent cells. More formally, (\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|) for all cells ((x, y)) and ((r, c)) if they are adjacent. Two cells are considered adjacent if they share a side. Find the minimum (\textit{beauty}) among all obtainable matrices. The first line contains a single integer (t) ((1 \le t \le 250)) — the number of test cases. The first line of each test case contains two integers (n) and (m) ((1 \le n, m \le 15)) — the number of rows and columns of (a), respectively. The next (n) lines, each containing (m) integers (a_{i,1}, a_{i,2}, \ldots, a_{i,m}) ((0 \le a_{i,j} < 2^{20})) — description of the matr