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>
const long double eps = 1e-9;
const long double pi = acos(-1.0);
using namespace std;
const int N = 300;
int a[N][N], used[N * N];
int n, m;
bool check(vector<int> r, vector<int> c) {
  vector<int> check_r, check_c;
  for (int i = 1; i <= n; i++) {
    int mx = 0;
    for (int j = 1; j <= m; j++) mx = max(mx, a[i][j]);
    check_r.push_back(mx);
  }
  for (int j = 1; j <= m; j++) {
    int mx = 0;
    for (int i = 1; i <= n; i++) mx = max(mx, a[i][j]);
    check_c.push_back(mx);
  }
  sort(check_r.begin(), check_r.end());
  sort(check_c.begin(), check_c.end());
  return (check_r == r && check_c == c);
}
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  vector<int> r, c;
  for (int i = 1; i <= n; i++) {
    int mx = 0;
    for (int j = 1; j <= m; j++) mx = max(mx, a[i][j]);
    r.push_back(mx);
  }
  for (int j = 1; j <= m; j++) {
    int mx = 0;
    for (int i = 1; i <= n; i++) mx = max(mx, a[i][j]);
    c.push_back(mx);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) a[i][j] = 0;
  sort(r.begin(), r.end());
  sort(c.begin(), c.end());
  vector<int> last_c, last_r;
  last_r = r, last_c = c;
  int posx, posy;
  posx = posy = 0;
  while (r.size() > 0 || c.size() > 0) {
    if (r.size() > 0 && c.size() > 0) {
      int R = r.back();
      int C = c.back();
      if (R == C) {
        posx++, posy++;
        a[posx][posy] = R;
        r.pop_back();
        c.pop_back();
        used[R] = used[C] = 1;
        continue;
      }
      if (C > R) {
        posy++;
        c.pop_back();
        a[posx][posy] = C;
        used[C] = 1;
      }
      if (R > C) {
        posx++;
        r.pop_back();
        a[posx][posy] = R;
        used[R] = 1;
      }
    } else if (r.size() > 0) {
      int R = r.back();
      posx++;
      r.pop_back();
      a[posx][posy] = R;
      used[R] = 1;
    } else {
      int C = c.back();
      posy++;
      c.pop_back();
      a[posx][posy] = C;
      used[C] = 1;
    }
  }
  vector<int> not_used;
  for (int i = 1; i <= n * m; i++)
    if (!used[i]) not_used.push_back(i);
  reverse(not_used.begin(), not_used.end());
  for (int i = n; i >= 1; i--) {
    for (int j = m; j >= 1; j--) {
      if (used[a[i][j]]) {
        for (int q = 1; q < i; q++)
          if (a[q][j] == 0) a[q][j] = not_used.back(), not_used.pop_back();
        for (int q = 1; q < j; q++)
          if (a[i][q] == 0) a[i][q] = not_used.back(), not_used.pop_back();
      }
    }
  }
  if (!check(last_r, last_c)) {
    cout << -1 << endl;
    return;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(10);
  int step;
  step = 1;
  for (int i = 1; i <= step; i++) {
    solve();
  }
}