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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline int msbp(int x) { return 31 - __builtin_clz(x); }
inline int msb(int x) { return 1 << msbp(x); }
const int INF = 0x3f3f3f3f;
const int N = 260;
int n, m, rm[N], cm[N], a[N][N], ans[N][N];
set<int> need;
void place(int r, int c) {
  ans[r][c] = *need.rbegin();
  need.erase(*need.rbegin());
}
bool verify() {
  set<int> orm, nrm, ocm, ncm;
  for (int i = 0; i < (n); ++i) {
    int omx = 0, nmx = 0;
    for (int j = 0; j < (m); ++j) {
      omx = max(omx, a[i][j]);
      nmx = max(nmx, ans[i][j]);
    }
    orm.insert(omx);
    nrm.insert(nmx);
  }
  for (int j = 0; j < (m); ++j) {
    int omx = 0, nmx = 0;
    for (int i = 0; i < (n); ++i) {
      omx = max(omx, a[i][j]);
      nmx = max(nmx, ans[i][j]);
    }
    ocm.insert(omx);
    ncm.insert(nmx);
  }
  if (orm != nrm || ocm != ncm)
    return false;
  else
    return true;
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = (1); i <= (n * m); ++i) need.insert(i);
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (m); ++j) {
      cin >> a[i][j];
      rm[i] = max(rm[i], a[i][j]);
      cm[j] = max(cm[j], a[i][j]);
    }
  }
  set<pair<int, int> > diag_nums;
  for (int i = 0; i < (n); ++i) {
    diag_nums.insert({rm[i], 0});
  }
  for (int j = 0; j < (m); ++j) {
    diag_nums.insert({cm[j], 1});
  }
  int r_id = -1, c_id = -1;
  vector<pair<int, int> > sp_ids;
  while (diag_nums.size()) {
    auto [cur, type] = *diag_nums.rbegin();
    diag_nums.erase(*diag_nums.rbegin());
    if (diag_nums.size() && diag_nums.rbegin()->first == cur) {
      diag_nums.erase(*diag_nums.rbegin());
      ans[++r_id][++c_id] = cur;
    } else {
      if (type == 0) {
        ans[++r_id][c_id] = cur;
      } else {
        ans[r_id][++c_id] = cur;
      }
    }
    need.erase(cur);
    sp_ids.push_back({r_id, c_id});
  }
  for (auto [r, c] : sp_ids) {
    for (int i = (r - 1); i >= (0); --i) {
      if (ans[i][c] == 0) place(i, c);
    }
    for (int j = (c - 1); j >= (0); --j) {
      if (ans[r][j] == 0) place(r, j);
    }
  }
  if (!verify()) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < (n); ++i) {
      for (int j = 0; j < (m); ++j) cout << ans[i][j] << ' ';
      cout << endl;
    }
  }
  return 0;
}