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;
const int INF = 2e9;
const long long ML = 4e18;
void solve() {
  int m, n;
  cin >> m >> n;
  vector<vector<int>> a(m, vector<int>(n));
  vector<int> x(m, 0), y(n, 0);
  for (int i = (0); i < (m); ++i) {
    for (int j = (0); j < (n); ++j) {
      cin >> a[i][j];
      x[i] = max(x[i], a[i][j]);
      y[j] = max(y[j], a[i][j]);
    }
  }
  vector<vector<int>> ans(m, vector<int>(n));
  set<int> sx = set<int>((x).begin(), (x).end()),
           sy = set<int>((y).begin(), (y).end());
  int px = 0, py = 0;
  vector<int> l(m), r(m), u(n), d(n);
  for (int k = (m * n); k >= (1); --k) {
    if (sx.count(k) && sy.count(k)) {
      ans[px][py] = k;
      l[px] = py - 1;
      r[px] = py + 1;
      u[py] = px - 1;
      d[py] = px + 1;
      px++, py++;
    } else if (sx.count(k)) {
      for (int j = (0); j < (py); ++j) {
        if (d[j] == px) {
          ans[px][j] = k;
          d[j]++;
          l[px] = j - 1;
          r[px] = j + 1;
          px++;
          break;
        }
      }
    } else if (sy.count(k)) {
      for (int j = (0); j < (px); ++j) {
        if (r[j] == py) {
          ans[j][py] = k;
          r[j]++;
          u[py] = j - 1;
          d[py] = j + 1;
          py++;
          break;
        }
      }
    } else {
      bool ok = false;
      for (int i = (0); i < (px); ++i) {
        if (l[i] != -1 && d[l[i]] == i) {
          ans[i][l[i]] = k;
          d[l[i]]++, l[i]--;
          ok = true;
          break;
        }
      }
      if (ok) continue;
      for (int i = (0); i < (px); ++i) {
        if (r[i] != n) {
          if (u[r[i]] == i) {
            ans[i][r[i]] = k;
            u[r[i]]--, r[i]++;
            break;
          }
          if (d[r[i]] == i) {
            ans[i][r[i]] = k;
            d[r[i]]++, r[i]++;
            break;
          }
        }
      }
    }
  }
  for (int i = (0); i < (m); ++i) {
    for (int j = (0); j < (n); ++j) {
      cout << ans[i][j] << " ";
    }
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}