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 = 1e9;
const long long inf64 = 1e18;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> a(n, vector<int>(m));
  vector<int> mx_row(n * m), mx_col(n * m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      x--;
      mx_row[i] = max(mx_row[i], x);
      mx_col[j] = max(mx_col[j], x);
      a[i][j] = x;
    }
  }
  vector<int> in_row(n * m), in_col(n * m);
  for (int i = 0; i < n; i++) {
    in_row[mx_row[i]] = 1;
  }
  for (int j = 0; j < m; j++) {
    in_col[mx_col[j]] = 1;
  }
  vector<vector<int>> mat;
  set<tuple<int, int, int>> poses;
  int nn = 0, mm = 0;
  auto upd = [&](int i, int j) {
    for (int x = 0; x <= 4; x++) poses.erase({x, i, j});
    int c = 0;
    if (i > 0 && mat[i - 1][j] != -1) c++;
    if (j > 0 && mat[i][j - 1] != -1) c++;
    if (i + 1 < nn && mat[i + 1][j] != -1) c++;
    if (j + 1 < mm && mat[i][j + 1] != -1) c++;
    poses.insert({c, i, j});
  };
  for (int x = n * m - 1; x >= 0; x--) {
    if (nn > n || mm > m) {
      cout << "-1\n";
      return 0;
    }
    if (in_row[x] && in_col[x]) {
      nn++;
      mm++;
      for (auto &r : mat) r.push_back(-1);
      mat.emplace_back(mm, -1);
      mat[nn - 1][mm - 1] = x;
      for (int i = 0; i < mm; i++) upd(nn - 1, i);
      for (int i = 0; i < nn; i++) upd(i, mm - 1);
    } else if (in_row[x]) {
      nn++;
      mat.emplace_back(mm, -1);
      for (int j = 0; j < mm; j++) {
        if (mat[nn - 2][j] != -1) {
          mat[nn - 1][j] = x;
          upd(nn - 2, j);
          break;
        }
      }
      for (int j = 0; j < mm; j++) upd(nn - 1, j);
    } else if (in_col[x]) {
      mm++;
      for (auto &r : mat) r.push_back(-1);
      for (int i = 0; i < nn; i++) {
        if (mat[i][mm - 2] != -1) {
          mat[i][mm - 1] = x;
          upd(i, mm - 2);
          break;
        }
      }
      for (int i = 0; i < nn; i++) upd(i, mm - 1);
    } else {
      while (!poses.empty()) {
        int i, j;
        tie(ignore, i, j) = *(--poses.end());
        if (mat[i][j] == -1) break;
        poses.erase(--poses.end());
      }
      if (poses.empty()) {
        cout << "-1\n";
        return 0;
      }
      int i, j;
      tie(ignore, i, j) = *(--poses.end());
      mat[i][j] = x;
      for (int di = -1; di <= 1; di++) {
        for (int dj = -1; dj <= 1; dj++) {
          if (di && dj) continue;
          int ti = i + di;
          int tj = j + dj;
          if (ti < 0 || ti >= nn || tj < 0 || tj >= mm) continue;
          upd(ti, tj);
        }
      }
    }
  }
  if (nn != n || mm != m) {
    cout << "-1\n";
    return 0;
  }
  for (auto r : mat) {
    for (auto x : r) cout << x + 1 << " ";
    cout << "\n";
  }
  return 0;
}