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>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(__attribute__((unused)) bool);
void precalc();
clock_t start;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
  }
  cout.flush();
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
template <typename T>
void make_unique(vector<T>& a) {
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
}
template <typename T>
void relax_min(T& cur, T val) {
  cur = min(cur, val);
}
template <typename T>
void relax_max(T& cur, T val) {
  cur = max(cur, val);
}
void precalc() {}
vector<vector<int>> g;
vector<int> order;
vector<char> used;
void dfs(int v) {
  if (used[v]) {
    return;
  }
  used[v] = true;
  for (int to : g[v]) {
    dfs(to);
  }
  order.push_back(v);
}
void solve(__attribute__((unused)) bool read) {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> a(n, vector<int>(m));
  vector<int> row_max(n), col_max(m);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> a[i][j];
      --a[i][j];
      relax_max(row_max[i], a[i][j]);
      relax_max(col_max[j], a[i][j]);
    }
  }
  vector<int> max_mask(n * m, 0);
  for (int i = 0; i < n; ++i) {
    max_mask[row_max[i]] |= 1;
  }
  for (int i = 0; i < m; ++i) {
    max_mask[col_max[i]] |= 2;
  }
  vector<int> row_max_index(n), col_max_index(m);
  vector<vector<int>> res(n, vector<int>(m, -1));
  int x = n - 1, y = m - 1;
  vector<int> free_vals;
  for (int val = n * m - 1; val >= 0;) {
    if (max_mask[val] == 3) {
      row_max_index[x] = y;
      col_max_index[y] = x;
      res[x][y] = val;
      --val;
      int cur_x = x - 1, cur_y = y - 1;
      while (val >= 0 && max_mask[val] != 3) {
        if (max_mask[val] == 1) {
          assert(cur_x >= 0);
          res[cur_x][y] = val;
          row_max_index[cur_x] = y;
          --cur_x;
        } else if (max_mask[val] == 2) {
          assert(cur_y >= 0);
          res[x][cur_y] = val;
          col_max_index[cur_y] = x;
          --cur_y;
        } else {
          free_vals.push_back(val);
        }
        --val;
      }
      x = cur_x;
      y = cur_y;
    }
  }
  g.clear();
  g.resize(n * m);
  auto get_num = [&](int x, int y) { return x * m + y; };
  auto add_edge = [&](int from_x, int from_y, int to_x, int to_y) {
    g[get_num(from_x, from_y)].push_back(get_num(to_x, to_y));
  };
  for (int i = 0; i < n; ++i) {
    int max_index = row_max_index[i];
    for (int j = max_index + 1; j < m; ++j) {
      add_edge(i, j - 1, i, j);
    }
    for (int j = max_index - 1; j >= 0; --j) {
      add_edge(i, j + 1, i, j);
    }
  }
  for (int i = 0; i < m; ++i) {
    int max_index = col_max_index[i];
    for (int j = max_index + 1; j < n; ++j) {
      add_edge(j - 1, i, j, i);
    }
    for (int j = max_index - 1; j >= 0; --j) {
      add_edge(j + 1, i, j, i);
    }
  }
  used.assign(n * m, false);
  for (int i = 0; i < n * m; ++i) {
    dfs(i);
  }
  sort(free_vals.begin(), free_vals.end());
  reverse(free_vals.begin(), free_vals.end());
  for (int key : order) {
    int x = key / m, y = key % m;
    if (res[x][y] != -1) {
      continue;
    }
    res[x][y] = free_vals.back();
    free_vals.pop_back();
  }
  for (auto& vec : res) {
    for (int x : vec) {
      cout << x + 1 << " ";
    }
    cout << "\n";
  }
}