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, i, j, a[255][255], r[255], c[255], f[62555], g[62555], b[255][255],
    col[255][255], sn, sm, vv[255][255];
vector<pair<int, int> > v;
void dfs(int x, int y) {
  if (b[x][y]) return;
  if (col[x][y]) return;
  col[x][y] = 1;
  dfs(x - 1, y);
  dfs(x, y + 1);
}
void gzz(int x, int y) {
  if (b[x][y] || vv[x][y]) return;
  if (col[x][y] && b[x - 1][y] && b[x][y + 1]) {
    vv[x][y] = 1;
    v.push_back(make_pair(x, y));
  }
  if (!col[x][y] && b[x + 1][y] && b[x][y - 1]) {
    vv[x][y] = 1;
    v.push_back(make_pair(x, y));
  }
}
void gz(int x, int y) {
  if (x > 1) gzz(x - 1, y);
  if (x < n) gzz(x + 1, y);
  if (y > 1) gzz(x, y - 1);
  if (y < m) gzz(x, y + 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for ((i) = 1; (i) <= (n); (i)++) {
    for ((j) = 1; (j) <= (m); (j)++) {
      scanf("%d", &a[i][j]);
      r[i] = max(r[i], a[i][j]);
      c[j] = max(c[j], a[i][j]);
    }
  }
  for ((i) = 1; (i) <= (n); (i)++) f[r[i]] = 1;
  for ((i) = 1; (i) <= (m); (i)++) g[c[i]] = 1;
  for (i = n * m; i; i--) {
    if (f[i] && g[i]) {
      sn++;
      sm++;
      b[sn][sm] = i;
      continue;
    }
    if (f[i]) {
      sn++;
      b[sn][sm] = i;
      continue;
    }
    if (g[i]) {
      sm++;
      b[sn][sm] = i;
      continue;
    }
  }
  if (!b[n][1]) {
    dfs(n, 1);
  }
  memset(b, 0, sizeof(b));
  sn = sm = 0;
  for (i = n * m; i; i--) {
    if (f[i] && g[i]) {
      sn++;
      sm++;
      b[sn][sm] = i;
      gz(sn, sm);
      continue;
    }
    if (f[i]) {
      sn++;
      b[sn][sm] = i;
      gz(sn, sm);
      continue;
    }
    if (g[i]) {
      sm++;
      b[sn][sm] = i;
      gz(sn, sm);
      continue;
    }
    int tx, ty;
    b[tx = v.back().first][ty = v.back().second] = i;
    v.pop_back();
    gz(tx, ty);
  }
  for ((i) = 1; (i) <= (n); (i)++) {
    for ((j) = 1; (j) <= (m); (j)++) {
      printf("%d%c", b[i][j], (j == m ? '\n' : ' '));
    }
  }
  return 0;
}