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;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int DIRS = 4;
const int DR[DIRS] = {-1, 0, +1, 0};
const int DC[DIRS] = {0, +1, 0, -1};
template <typename T>
void output_vector(const vector<T> &v, bool add_one = false, int start = -1,
                   int end = -1) {
  if (start < 0) start = 0;
  if (end < 0) end = int(v.size());
  for (int i = start; i < end; i++)
    cout << v[i] + (add_one ? 1 : 0) << (i < end - 1 ? ' ' : '\n');
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, M;
  cin >> N >> M;
  vector<vector<int>> A(N, vector<int>(M));
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) cin >> A[i][j];
  vector<int> row_max(N, 0);
  vector<int> col_max(M, 0);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) {
      row_max[i] = max(row_max[i], A[i][j]);
      col_max[j] = max(col_max[j], A[i][j]);
    }
  sort(row_max.begin(), row_max.end());
  sort(col_max.begin(), col_max.end());
  vector<vector<int>> neighbors(N, vector<int>(M, 0));
  vector<set<pair<int, int>>> neighbor_sets(DIRS + 1);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) neighbor_sets[0].insert({i, j});
  vector<vector<int>> solution(N, vector<int>(M, -1));
  vector<bool> done(N * M + 1, false);
  auto &&place = [&](int r, int c, int number) {
    ;
    assert(solution[r][c] < 0);
    neighbor_sets[neighbors[r][c]].erase({r, c});
    solution[r][c] = number;
    done[number] = true;
    for (int dir = 0; dir < DIRS; dir++) {
      int nr = r + DR[dir];
      int nc = c + DC[dir];
      if (0 <= nr && nr < N && 0 <= nc && nc < M) {
        neighbor_sets[neighbors[nr][nc]].erase({nr, nc});
        neighbors[nr][nc]++;
        if (solution[nr][nc] < 0)
          neighbor_sets[neighbors[nr][nc]].insert({nr, nc});
      }
    }
  };
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++)
      if (row_max[i] == col_max[j]) place(i, j, row_max[i]);
  for (int number = N * M; number > 0; number--)
    if (!done[number]) {
      int row = -1, col = -1;
      for (int i = 0; i < N; i++)
        if (row_max[i] == number) row = i;
      for (int j = 0; j < M; j++)
        if (col_max[j] == number) col = j;
      if (row >= 0) {
        int best_col = -1;
        for (int j = 0; j < M; j++)
          if (solution[row][j] < 0 && col_max[j] >= number &&
              (best_col < 0 || neighbors[row][j] > neighbors[row][best_col]))
            best_col = j;
        assert(best_col >= 0);
        place(row, best_col, number);
      } else if (col >= 0) {
        int best_row = -1;
        for (int i = 0; i < N; i++)
          if (solution[i][col] < 0 && row_max[i] >= number &&
              (best_row < 0 || neighbors[i][col] > neighbors[best_row][col]))
            best_row = i;
        assert(best_row >= 0);
        place(best_row, col, number);
      } else {
        for (int neigh = DIRS; neigh >= 0; neigh--)
          for (pair<int, int> p : neighbor_sets[neigh])
            if (row_max[p.first] >= number && col_max[p.second] >= number) {
              neighbor_sets[neigh].erase(p);
              place(p.first, p.second, number);
              goto finish;
            }
      finish:;
      }
    }
  for (auto &row : solution) output_vector(row);
}