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;
using pii = pair<int, int>;
void print(vector<vector<int>> &a) {
  for (auto &j : a) {
    for (int &i : j) cout << i << ' ';
    cout << '\n';
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> A(n, vector<int>(m));
  for (auto &i : A)
    for (int &j : i) cin >> j;
  vector<int> mx_vert(n), mx_hor(m);
  for (int i = 0; i < n; i++) {
    int ans = -1;
    for (int j = 0; j < m; j++) {
      ans = max(ans, A[i][j]);
    }
    mx_vert[i] = ans;
  }
  for (int j = 0; j < m; j++) {
    int ans = -1;
    for (int i = 0; i < n; i++) {
      ans = max(ans, A[i][j]);
    }
    mx_hor[j] = ans;
  }
  sort(mx_vert.begin(), mx_vert.end());
  sort(mx_hor.begin(), mx_hor.end());
  int mega = m * n;
  vector<int> new_diag, new_vert, new_hor;
  set_intersection(mx_vert.begin(), mx_vert.end(), mx_hor.begin(), mx_hor.end(),
                   back_inserter(new_diag));
  set_difference(mx_vert.begin(), mx_vert.end(), new_diag.begin(),
                 new_diag.end(), back_inserter(new_vert));
  set_difference(mx_hor.begin(), mx_hor.end(), new_diag.begin(), new_diag.end(),
                 back_inserter(new_hor));
  vector<vector<int>> ANS(n, vector<int>(m));
  int center = new_diag.size() - 1;
  vector<pii> loc(n * m + 69);
  for (int i = 0; i <= center; i++) {
    ANS[i][i] = new_diag[i];
    loc[new_diag[i]] = {i, i};
  }
  for (int i = n - 1, j = 0; i > center; i--, j++) {
    ANS[i][center] = new_vert[j];
    loc[new_vert[j]] = {i, center};
  }
  for (int i = m - 1, j = 0; i > center; i--, j++) {
    ANS[center][i] = new_hor[j];
    loc[new_hor[j]] = {center, i};
  }
  priority_queue<int, vector<int>, greater<int>> filling;
  vector<int> filler;
  for (int i = 1; i <= n * m; i++) {
    if (!binary_search(new_diag.begin(), new_diag.end(), i) &&
        !binary_search(new_hor.begin(), new_hor.end(), i) &&
        !binary_search(new_vert.begin(), new_vert.end(), i))
      filler.push_back(i);
  }
  for (int i : new_diag) filling.push(i);
  for (int i : new_hor) filling.push(i);
  for (int i : new_vert) filling.push(i);
  int fills = 0;
  while (!filling.empty()) {
    int now = filling.top();
    filling.pop();
    if (now == n * m) break;
    auto [x, y] = loc[now];
    if (x == y) {
      for (int i = n - 1; i > x; i--) {
        if (ANS[i][y])
          continue;
        else
          ANS[i][y] = filler[fills++];
      }
      for (int i = m - 1; i > y; i--) {
        if (ANS[x][i])
          continue;
        else
          ANS[x][i] = filler[fills++];
      }
    } else if (x == center) {
      for (int i = 0; i < center; i++) {
        if (ANS[i][y])
          continue;
        else
          ANS[i][y] = filler[fills++];
      }
      for (int i = n - 1; i > center; i--) {
        if (ANS[i][y])
          continue;
        else
          ANS[i][y] = filler[fills++];
      }
    } else if (y == center) {
      for (int i = 0; i < center; i++) {
        if (ANS[x][i])
          continue;
        else
          ANS[x][i] = filler[fills++];
      }
      for (int i = m - 1; i > center; i--) {
        if (ANS[x][i])
          continue;
        else
          ANS[x][i] = filler[fills++];
      }
    }
  }
  print(ANS);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int owo = 1;
  while (owo--) solve();
}