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 \}) 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[253][253];
int b[253][253];
int p[253 * 253];
int px[253], py[253];
int vx[253], vy[253];
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
bool valid(int i, int j) {
  if (i < 0 || i >= n || j < 0 || j >= m) return false;
  if (b[i][j]) return false;
  int k = 0;
  if (i + 1 < n && j - 1 >= 0 && b[i + 1][j] > k && b[i][j - 1] > k)
    return true;
  if (i - 1 >= 0 && j + 1 < m && b[i - 1][j] > k && b[i][j + 1] > k)
    return true;
  return false;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      px[i] = max(px[i], a[i][j]);
      py[j] = max(py[j], a[i][j]);
    }
  for (int i = 0; i < n; i++) p[px[i]] ^= 1;
  for (int j = 0; j < m; j++) p[py[j]] ^= 2;
  sort(px, px + n);
  reverse(px, px + n);
  sort(py, py + m);
  reverse(py, py + m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++)
      if (py[j] >= px[i]) vx[i]++;
    vx[i]--;
    b[i][vx[i]] = px[i];
  }
  for (int j = 0; j < m; j++) {
    for (int i = 0; i < n; i++)
      if (px[i] >= py[j]) vy[j]++;
    vy[j]--;
    b[vy[j]][j] = py[j];
  }
  set<pair<int, int> > A;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (valid(i, j)) A.insert(make_pair(i, j));
  for (int k = n * m; k >= 1; k--)
    if (!p[k]) {
      bool F = false;
      for (auto p : A) {
        int i = p.first;
        int j = p.second;
        if (i + 1 < n && j - 1 >= 0 && b[i + 1][j] > k && b[i][j - 1] > k ||
            i - 1 >= 0 && j + 1 < m && b[i - 1][j] > k && b[i][j + 1] > k) {
          b[i][j] = k;
          F = true;
          A.erase(make_pair(i, j));
          for (int d = 0; d < 4; d++) {
            if (valid(i + dx[d], j + dy[d]))
              A.insert(make_pair(i + dx[d], j + dy[d]));
          }
          break;
        }
      }
      assert(F);
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      printf("%d%c", b[i][j], j == m - 1 ? '\n' : ' ');
  return 0;
}