Rearrange
Koa the Koala has a matrix $$$A$$$ of $$$n$$$ rows and $$$m$$$ columns. Elements of this matrix are distinct integers from $$$1$$$ to $$$n \cdot m$$$ (each number from $$$1$$$ to $$$n \cdot m$$$ appears exactly once in the matrix).For any matrix $$$M$$$ of $$$n$$$ rows and $$$m$$$ columns let's define the following:  The $$$i$$$-th row of $$$M$$$ is defined as $$$R_i(M) = [ M_{i1}, M_{i2}, \ldots, M_{im} ]$$$ for all $$$i$$$ ($$$1 \le i \le n$$$).  The $$$j$$$-th column of $$$M$$$ is defined as $$$C_j(M) = [ M_{1j}, M_{2j}, \ldots, M_{nj} ]$$$ for all $$$j$$$ ($$$1 \le j \le m$$$). Koa defines $$$S(A) = (X, Y)$$$ as the spectrum of $$$A$$$, where $$$X$$$ is the set of the maximum values in rows of $$$A$$$ and $$$Y$$$ is the set of the maximum values in columns of $$$A$$$.More formally:  $$$X = \{ \max(R_1(A)), \max(R_2(A)), \ldots, \max(R_n(A)) \}$$$  $$$Y = \{ \max(C_1(A)), \max(C_2(A)), \ldots, \max(C_m(A)) \}$$$Koa asks you to find some matrix $$$A'$$$ of $$$n$$$ rows and $$$m$$$ columns, such that each number from $$$1$$$ to $$$n \cdot m$$$ appears exactly once in the matrix, and the following conditions hold:  $$$S(A') = S(A)$$$  $$$R_i(A')$$$ is bitonic for all $$$i$$$ ($$$1 \le i \le n$$$)  $$$C_j(A')$$$ is bitonic for all $$$j$$$ ($$$1 \le j \le m$$$)  An array $$$t$$$ ($$$t_1, t_2, \ldots, t_k$$$) is called bitonic if it first increases and then decreases. More formally: $$$t$$$ is bitonic if there exists some position $$$p$$$ ($$$1 \le p \le k$$$) such that: $$$t_1 < t_2 < \ldots < t_p > t_{p+1} > \ldots > t_k$$$.Help Koa to find such matrix or to determine that it doesn't exist.
Let's analyze the first sample:For matrix $$$A$$$ we have:   Rows:   $$$R_1(A) = [3, 5, 6]; \max(R_1(A)) = 6$$$  $$$R_2(A) = [1, 7, 9]; \max(R_2(A)) = 9$$$  $$$R_3(A) = [4, 8, 2]; \max(R_3(A)) = 8$$$   Columns:   $$$C_1(A) = [3, 1, 4]; \max(C_1(A)) = 4$$$  $$$C_2(A) = [5, 7, 8]; \max(C_2(A)) = 8$$$  $$$C_3(A) = [6, 9, 2]; \max(C_3(A)) = 9$$$    $$$X = \{ \max(R_1(A)), \max(R_2(A)), \max(R_3(A)) \} = \{ 6, 9, 8 \}$$$  $$$Y = \{ \max(C_1(A)), \max(C_2(A)), \max(C_3(A)) \} = \{ 4, 8, 9 \}$$$  So $$$S(A) = (X, Y) = (\{ 6, 9, 8 \}, \{ 4, 8, 9 \})$$$ For matrix $$$A'$$$ we have:   Rows:   $$$R_1(A') = [9, 5, 1]; \max(R_1(A')) = 9$$$  $$$R_2(A') = [7, 8, 2]; \max(R_2(A')) = 8$$$  $$$R_3(A') = [3, 6, 4]; \max(R_3(A')) = 6$$$   Columns:   $$$C_1(A') = [9, 7, 3]; \max(C_1(A')) = 9$$$  $$$C_2(A') = [5, 8, 6]; \max(C_2(A')) = 8$$$  $$$C_3(A') = [1, 2, 4]; \max(C_3(A')) = 4$$$    Note that each of this arrays are bitonic.  $$$X = \{ \max(R_1(A')), \max(R_2(A')), \max(R_3(A')) \} = \{ 9, 8, 6 \}$$$  $$$Y = \{ \max(C_1(A')), \max(C_2(A')), \max(C_3(A')) \} = \{ 9, 8, 4 \}$$$  So $$$S(A') = (X, Y) = (\{ 9, 8, 6 \}, \{ 9, 8, 4 \})$$$ 
The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$)  — the number of rows and columns of $$$A$$$.Each of the ollowing $$$n$$$ lines contains $$$m$$$ integers. The $$$j$$$-th integer in the $$$i$$$-th line denotes element $$$A_{ij}$$$ ($$$1 \le A_{ij} \le n \cdot m$$$) of matrix $$$A$$$. It is guaranteed that every number from $$$1$$$ to $$$n \cdot m$$$ appears exactly once among elements of the matrix.
If such matrix doesn't exist, print $$$-1$$$ on a single line.Otherwise, the output must consist of $$$n$$$ lines, each one consisting of $$$m$$$ space separated integers  — a description of $$$A'$$$.The $$$j$$$-th number in the $$$i$$$-th line represents the element $$$A'_{ij}$$$.Every integer from $$$1$$$ to $$$n \cdot m$$$ should appear exactly once in $$$A'$$$, every row and column in $$$A'$$$ must be bitonic and $$$S(A) = S(A')$$$ must hold.If there are many answers print any.
(n, m) = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(n)]
line = [0] * (n * m + 1)
col = [0] * (n * m + 1)
ans = [[0] * m for _ in range(n)]
q = [0] * (n * m + 1)
for i in range(n):
    ma = max(a[i])
    line[ma] = 1
for j in range(m):
    ma = max((a[i][j] for i in range(n)))
    col[ma] = 1
(X, Y, st, ed) = (0, 0, 0, 0)
for i in range(n * m, 0, -1):
    X += line[i]
    Y += col[i]
    if line[i] or col[i]:
        ans[X - 1][Y - 1] = i
    else:
        ans[q[st][0] - 1][q[st][1] - 1] = i
        st += 1
    if line[i]:
        for j in range(Y - 1, 0, -1):
            q[ed] = (X, j)
            ed += 1
    if col[i]:
        for j in range(X - 1, 0, -1):
            q[ed] = (j, Y)
            ed += 1
for row in ans:
    print(*row)
(n, m) = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(n)]
line = [0] * (n * m + 1)
col = [0] * (n * m + 1)
ans = [[0] * m for _ in range(n)]
q = [0] * (n * m + 1)
for i in range(n):
    ma = max(a[i])
    line[ma] = 1
for j in range(m):
    ma = max((a[i][j] for i in range(n)))
    col[ma] = 1
(X, Y, st, ed) = (0, 0, 0, 0)
for i in range(n * m, 0, -1):
    X += line[i]
    Y += col[i]
    if line[i] or col[i]:
        ans[X - 1][Y - 1] = i
    else:
        ans[q[st][0] - 1][q[st][1] - 1] = i
        st += 1
    if line[i]:
        for j in range(Y - 1, 0, -1):
            q[ed] = (X, j)
            ed += 1
    if col[i]:
        for j in range(X - 1, 0, -1):
            q[ed] = (j, Y)
            ed += 1
for row in ans:
    print(*row)
(n, m) = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(n)]
line = [0] * (n * m + 1)
col = [0] * (n * m + 1)
ans = [[0] * m for _ in range(n)]
q = [0] * (n * m + 1)
for i in range(n):
    ma = max(a[i])
    line[ma] = 1
for j in range(m):
    ma = max((a[i][j] for i in range(n)))
    col[ma] = 1
(X, Y, st, ed) = (0, 0, 0, 0)
for i in range(n * m, 0, -1):
    X += line[i]
    Y += col[i]
    if line[i] or col[i]:
        ans[X - 1][Y - 1] = i
    else:
        ans[q[st][0] - 1][q[st][1] - 1] = i
        st += 1
    if line[i]:
        for j in range(Y - 1, 0, -1):
            q[ed] = (X, j)
            ed += 1
    if col[i]:
        for j in range(X - 1, 0, -1):
            q[ed] = (j, Y)
            ed += 1
for row in ans:
    print(*row)