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, typename U>
bool ckmin(T &a, U b) {
  return b < a ? a = b, true : false;
}
template <typename T, typename U>
bool ckmax(T &a, U b) {
  return b > a ? a = b, true : false;
}
int n, m;
set<int, greater<int>> rem;
vector<int> x, y, ax, ay;
vector<vector<int>> v, a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  x.assign(n, 0);
  y.assign(m, 0);
  ax.assign(n, 0);
  ay.assign(m, 0);
  v.assign(n, vector<int>(m, 0));
  a.assign(n, vector<int>(m, 0));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> v[i][j];
    }
  }
  for (int i = 0; i < n * m; ++i) {
    rem.insert(i + 1);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      ckmax(x[i], v[i][j]);
    }
  }
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      ckmax(y[i], v[j][i]);
    }
  }
  set<int, greater<int>> xx, yy;
  for (auto i : x) xx.insert(i);
  for (auto i : y) yy.insert(i);
  for (auto i : x) {
    if (rem.find(i) != rem.end()) rem.erase(i);
  }
  for (auto j : y) {
    if (rem.find(j) != rem.end()) rem.erase(j);
  }
  int r = -1, c = -1;
  while ((int)(xx).size() && (int)(yy).size()) {
    auto it1 = xx.begin(), it2 = yy.begin();
    if (*it1 == *it2) {
      r++, c++;
      a[r][c] = *it1;
      xx.erase(it1);
      yy.erase(it2);
    } else if (*it1 > *it2) {
      r++;
      a[r][c] = *it1;
      xx.erase(it1);
    } else {
      c++;
      a[r][c] = *it2;
      yy.erase(it2);
    }
  }
  while ((int)(yy).size()) {
    c++;
    a[r][c] = *yy.begin();
    yy.erase(yy.begin());
  }
  while ((int)(xx).size()) {
    r++;
    a[r][c] = *xx.begin();
    xx.erase(xx.begin());
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      ckmax(ax[i], a[i][j]);
    }
  }
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      ckmax(ay[i], a[j][i]);
    }
  }
  sort(x.begin(), x.end());
  sort(y.begin(), y.end());
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (!a[i][j]) continue;
      queue<pair<int, int>> q;
      q.push(make_pair(i - 1, 0));
      q.push(make_pair(j - 1, 1));
      while ((int)(q).size()) {
        pair<int, int> x = q.front();
        q.pop();
        auto it = rem.lower_bound(a[i][j]);
        if (x.second) {
          if (x.first < 0) continue;
          if (!a[i][x.first]) {
            assert(it != rem.end());
            a[i][x.first] = *it;
            rem.erase(it);
          }
          q.push(make_pair(x.first - 1, 1));
        } else {
          if (x.first < 0) continue;
          if (!a[x.first][j]) {
            assert(it != rem.end());
            a[x.first][j] = *it;
            rem.erase(it);
          }
          q.push(make_pair(x.first - 1, 0));
        }
      }
    }
  }
  sort(ay.begin(), ay.end());
  sort(ax.begin(), ax.end());
  if (ay == y && ax == x) {
    for (auto i : a) {
      for (auto j : i) {
        cout << j << " ";
      }
      cout << "\n";
    }
  } else
    cout << "-1";
  return 0;
}