#include <bits/stdc++.h>
using namespace std;
bool inside(int x, int y, int n, int m) {
  return (0 <= x && x < n && 0 <= y && y < m);
}
void update_n(int x, int y, int n, int m, vector<vector<int>>& neighbors) {
  if (inside(x + 1, y, n, m)) {
    neighbors[x + 1][y]++;
  }
  if (inside(x - 1, y, n, m)) {
    neighbors[x - 1][y]++;
  }
  if (inside(x, y + 1, n, m)) {
    neighbors[x][y + 1]++;
  }
  if (inside(x, y - 1, n, m)) {
    neighbors[x][y - 1]++;
  }
}
int max_n(int x, int y, int n, int m, vector<vector<int>>& b) {
  int max_val = 0;
  if (inside(x + 1, y, n, m)) {
    max_val = max(max_val, b[x + 1][y]);
  }
  if (inside(x - 1, y, n, m)) {
    max_val = max(max_val, b[x - 1][y]);
  }
  if (inside(x, y + 1, n, m)) {
    max_val = max(max_val, b[x][y + 1]);
  }
  if (inside(x, y - 1, n, m)) {
    max_val = max(max_val, b[x][y - 1]);
  }
  return max_val;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> a(n, vector<int>(m, 0));
  vector<vector<int>> b(n, vector<int>(m, 0));
  vector<vector<int>> neighbors(n, vector<int>(m, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  priority_queue<int> row_pq;
  priority_queue<int> col_pq;
  unordered_set<int> used_nums;
  for (int i = 0; i < n; i++) {
    int max_row = 0;
    for (int j = 0; j < m; j++) {
      max_row = max(max_row, a[i][j]);
    }
    row_pq.push(max_row);
    used_nums.insert(max_row);
  }
  for (int i = 0; i < m; i++) {
    int max_col = 0;
    for (int j = 0; j < n; j++) {
      max_col = max(max_col, a[j][i]);
    }
    col_pq.push(max_col);
    used_nums.insert(max_col);
  }
  int x = -1;
  int y = -1;
  while (!row_pq.empty() || !col_pq.empty()) {
    int q = 0;
    if (row_pq.empty()) {
      q = col_pq.top();
      col_pq.pop();
      y++;
    } else if (col_pq.empty()) {
      q = row_pq.top();
      row_pq.pop();
      x++;
    } else {
      int r = row_pq.top();
      int c = col_pq.top();
      if (r == c) {
        row_pq.pop();
        col_pq.pop();
        x++;
        y++;
        q = r;
      } else if (r < c) {
        q = c;
        col_pq.pop();
        y++;
      } else {
        q = r;
        row_pq.pop();
        x++;
      }
    }
    b[x][y] = q;
    update_n(x, y, n, m, neighbors);
  }
  priority_queue<pair<int, pair<int, int>>> pq;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
  }
  int num = n * m;
  while (!pq.empty()) {
    auto p = pq.top();
    x = p.second.first;
    y = p.second.second;
    pq.pop();
    while (used_nums.count(num) > 0) {
      num--;
    }
    b[x][y] = num;
    used_nums.insert(num);
    update_n(x, y, n, m, neighbors);
    if (inside(x + 1, y, n, m)) {
      int i = x + 1;
      int j = y;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
    if (inside(x - 1, y, n, m)) {
      int i = x - 1;
      int j = y;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
    if (inside(x, y + 1, n, m)) {
      int i = x;
      int j = y + 1;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
    if (inside(x, y - 1, n, m)) {
      int i = x;
      int j = y - 1;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cout << b[i][j] << " ";
    }
    cout << endl;
  }
}

### Prompt

Your task is to create a Cpp solution to the following 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 \}) 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool inside(int x, int y, int n, int m) {
  return (0 <= x && x < n && 0 <= y && y < m);
}
void update_n(int x, int y, int n, int m, vector<vector<int>>& neighbors) {
  if (inside(x + 1, y, n, m)) {
    neighbors[x + 1][y]++;
  }
  if (inside(x - 1, y, n, m)) {
    neighbors[x - 1][y]++;
  }
  if (inside(x, y + 1, n, m)) {
    neighbors[x][y + 1]++;
  }
  if (inside(x, y - 1, n, m)) {
    neighbors[x][y - 1]++;
  }
}
int max_n(int x, int y, int n, int m, vector<vector<int>>& b) {
  int max_val = 0;
  if (inside(x + 1, y, n, m)) {
    max_val = max(max_val, b[x + 1][y]);
  }
  if (inside(x - 1, y, n, m)) {
    max_val = max(max_val, b[x - 1][y]);
  }
  if (inside(x, y + 1, n, m)) {
    max_val = max(max_val, b[x][y + 1]);
  }
  if (inside(x, y - 1, n, m)) {
    max_val = max(max_val, b[x][y - 1]);
  }
  return max_val;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> a(n, vector<int>(m, 0));
  vector<vector<int>> b(n, vector<int>(m, 0));
  vector<vector<int>> neighbors(n, vector<int>(m, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  priority_queue<int> row_pq;
  priority_queue<int> col_pq;
  unordered_set<int> used_nums;
  for (int i = 0; i < n; i++) {
    int max_row = 0;
    for (int j = 0; j < m; j++) {
      max_row = max(max_row, a[i][j]);
    }
    row_pq.push(max_row);
    used_nums.insert(max_row);
  }
  for (int i = 0; i < m; i++) {
    int max_col = 0;
    for (int j = 0; j < n; j++) {
      max_col = max(max_col, a[j][i]);
    }
    col_pq.push(max_col);
    used_nums.insert(max_col);
  }
  int x = -1;
  int y = -1;
  while (!row_pq.empty() || !col_pq.empty()) {
    int q = 0;
    if (row_pq.empty()) {
      q = col_pq.top();
      col_pq.pop();
      y++;
    } else if (col_pq.empty()) {
      q = row_pq.top();
      row_pq.pop();
      x++;
    } else {
      int r = row_pq.top();
      int c = col_pq.top();
      if (r == c) {
        row_pq.pop();
        col_pq.pop();
        x++;
        y++;
        q = r;
      } else if (r < c) {
        q = c;
        col_pq.pop();
        y++;
      } else {
        q = r;
        row_pq.pop();
        x++;
      }
    }
    b[x][y] = q;
    update_n(x, y, n, m, neighbors);
  }
  priority_queue<pair<int, pair<int, int>>> pq;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
  }
  int num = n * m;
  while (!pq.empty()) {
    auto p = pq.top();
    x = p.second.first;
    y = p.second.second;
    pq.pop();
    while (used_nums.count(num) > 0) {
      num--;
    }
    b[x][y] = num;
    used_nums.insert(num);
    update_n(x, y, n, m, neighbors);
    if (inside(x + 1, y, n, m)) {
      int i = x + 1;
      int j = y;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
    if (inside(x - 1, y, n, m)) {
      int i = x - 1;
      int j = y;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
    if (inside(x, y + 1, n, m)) {
      int i = x;
      int j = y + 1;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
    if (inside(x, y - 1, n, m)) {
      int i = x;
      int j = y - 1;
      if (b[i][j] == 0 && neighbors[i][j] == 2) {
        pq.push({-min(i, j), {i, j}});
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cout << b[i][j] << " ";
    }
    cout << endl;
  }
}

```