#include <bits/stdc++.h>
using namespace std;
int n, m, a[505][505], i, j, k, visx[505], visy[505], l, p[500005], mxx[505],
    mxy[505], tot, ans[505][505], lx, ly, x, y;
int Nex() {
  while (p[l]) ++l;
  p[l] = 1;
  return l;
}
int main() {
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) scanf("%d", &a[i][j]);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) mxx[i] = max(mxx[i], a[i][j]);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) mxy[j] = max(mxy[j], a[i][j]);
  sort(mxx + 1, mxx + 1 + n);
  sort(mxy + 1, mxy + 1 + m);
  mxx[n + 1] = mxy[m + 1] = 1 << 30;
  lx = ly = 1;
  x = n, y = m;
  while (lx <= n || ly <= m)
    if (mxx[lx] < mxy[ly])
      ans[x--][y] = mxx[lx++];
    else if (mxx[lx] > mxy[ly])
      ans[x][y--] = mxy[ly++];
    else {
      ans[x--][y--] = mxx[lx++];
      ++ly;
    }
  l = tot;
  for (i = 1; i <= n; ++i) {
    p[mxx[i]] = 1;
    if (!visx[i]) ans[++l][tot] = mxx[i];
  }
  l = tot;
  for (i = 1; i <= m; ++i) {
    p[mxy[i]] = 1;
    if (!visy[i]) ans[tot][++l] = mxy[i];
  }
  l = 1;
  for (i = 1; i <= n * m; ++i)
    if (p[i]) {
      int x = 0, y = 0;
      for (j = 1; j <= n; ++j)
        for (k = 1; k <= m; ++k)
          if (ans[j][k] == i) {
            x = j;
            y = k;
            break;
          }
      for (j = x - 1; !ans[j][y] && j >= 1; --j)
        ;
      for (++j; j < x; ++j)
        if (!ans[j][y]) ans[j][y] = Nex();
      for (j = x + 1; !ans[j][y] && j <= n; ++j)
        ;
      for (--j; j > x; --j)
        if (!ans[j][y]) ans[j][y] = Nex();
      for (j = y - 1; !ans[x][j] && j >= 1; --j)
        ;
      for (++j; j < y; ++j)
        if (!ans[x][j]) ans[x][j] = Nex();
      for (j = y + 1; !ans[x][j] && j <= m; ++j)
        ;
      for (--j; j > y; --j)
        if (!ans[x][j]) ans[x][j] = Nex();
    }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) printf("%d ", ans[i][j]);
    printf("\n");
  }
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

Koa the Koala has a matrix A of n rows and m columns. Elements of this matrix are distinct integers from 1 to n ⋅ m (each number from 1 to n ⋅ 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}, …, M_{im} ] for all i (1 ≤ i ≤ n). 
  * The j-th column of M is defined as C_j(M) = [ M_{1j}, M_{2j}, …, M_{nj} ] for all j (1 ≤ j ≤ 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)), …, max(R_n(A)) \} 
  * Y = \{ max(C_1(A)), max(C_2(A)), …, 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 ⋅ 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 ≤ i ≤ n) 
  * C_j(A') is bitonic for all j (1 ≤ j ≤ m) 

An array t (t_1, t_2, …, t_k) is called bitonic if it first increases and then decreases.

More formally: t is bitonic if there exists some position p (1 ≤ p ≤ k) such that: t_1 < t_2 < … < t_p > t_{p+1} > … > t_k.

Help Koa to find such matrix or to determine that it doesn't exist.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 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 ≤ A_{ij} ≤ n ⋅ m) of matrix A. It is guaranteed that every number from 1 to n ⋅ m appears exactly once among elements of the matrix.

Output

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 ⋅ 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.

Examples

Input


3 3
3 5 6
1 7 9
4 8 2


Output


9 5 1
7 8 2
3 6 4


Input


2 2
4 1
3 2


Output


4 1
3 2


Input


3 4
12 10 8 6
3 4 5 7
2 11 9 1


Output


12 8 6 1
10 11 9 2
3 4 5 7

Note

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 \}) 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, a[505][505], i, j, k, visx[505], visy[505], l, p[500005], mxx[505],
    mxy[505], tot, ans[505][505], lx, ly, x, y;
int Nex() {
  while (p[l]) ++l;
  p[l] = 1;
  return l;
}
int main() {
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) scanf("%d", &a[i][j]);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) mxx[i] = max(mxx[i], a[i][j]);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) mxy[j] = max(mxy[j], a[i][j]);
  sort(mxx + 1, mxx + 1 + n);
  sort(mxy + 1, mxy + 1 + m);
  mxx[n + 1] = mxy[m + 1] = 1 << 30;
  lx = ly = 1;
  x = n, y = m;
  while (lx <= n || ly <= m)
    if (mxx[lx] < mxy[ly])
      ans[x--][y] = mxx[lx++];
    else if (mxx[lx] > mxy[ly])
      ans[x][y--] = mxy[ly++];
    else {
      ans[x--][y--] = mxx[lx++];
      ++ly;
    }
  l = tot;
  for (i = 1; i <= n; ++i) {
    p[mxx[i]] = 1;
    if (!visx[i]) ans[++l][tot] = mxx[i];
  }
  l = tot;
  for (i = 1; i <= m; ++i) {
    p[mxy[i]] = 1;
    if (!visy[i]) ans[tot][++l] = mxy[i];
  }
  l = 1;
  for (i = 1; i <= n * m; ++i)
    if (p[i]) {
      int x = 0, y = 0;
      for (j = 1; j <= n; ++j)
        for (k = 1; k <= m; ++k)
          if (ans[j][k] == i) {
            x = j;
            y = k;
            break;
          }
      for (j = x - 1; !ans[j][y] && j >= 1; --j)
        ;
      for (++j; j < x; ++j)
        if (!ans[j][y]) ans[j][y] = Nex();
      for (j = x + 1; !ans[j][y] && j <= n; ++j)
        ;
      for (--j; j > x; --j)
        if (!ans[j][y]) ans[j][y] = Nex();
      for (j = y - 1; !ans[x][j] && j >= 1; --j)
        ;
      for (++j; j < y; ++j)
        if (!ans[x][j]) ans[x][j] = Nex();
      for (j = y + 1; !ans[x][j] && j <= m; ++j)
        ;
      for (--j; j > y; --j)
        if (!ans[x][j]) ans[x][j] = Nex();
    }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) printf("%d ", ans[i][j]);
    printf("\n");
  }
}

```