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;
const __int128 ONE = 1;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int mn = 534;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
int a[mn][mn], r[mn], c[mn], res[mn][mn];
pair<int, int> pos[mn * mn];
bool u[mn * mn], vis[mn][mn];
vector<pair<int, int> > g[mn][mn];
struct comp {
  bool operator()(pair<int, int> a, pair<int, int> b) {
    return min(r[a.first], c[a.second]) < min(r[b.first], c[b.second]);
  }
};
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      r[i] = max(r[i], a[i][j]);
      c[j] = max(c[j], a[i][j]);
    }
  sort(r, r + n, greater<int>());
  sort(c, c + m, greater<int>());
  memset(a, 0, sizeof(a));
  int ci = 0, cj = 0;
  for (int i = 0, j = 0; i < n || j < m;) {
    if (1) {
      if (r[i] > c[j]) {
        i++;
        u[r[i - 1]] = 1;
        vis[i - 1][j - 1] = 1;
        a[i - 1][j - 1] = r[i - 1];
        pos[r[i - 1]] = {i - 1, j - 1};
      } else {
        if (r[i] == c[j]) i++;
        j++;
        u[c[j - 1]] = 1;
        vis[i - 1][j - 1] = 1;
        a[i - 1][j - 1] = c[j - 1];
        pos[c[j - 1]] = {i - 1, j - 1};
      }
      ci = i - 1, cj = j - 1;
      for (int ii = ci; ii + 1 < n; ii++)
        g[ii][cj].push_back({ii + 1, cj}), res[ii + 1][cj]++;
      for (int ii = ci; ii - 1 >= 0; ii--)
        g[ii][cj].push_back({ii - 1, cj}), res[ii - 1][cj]++;
      for (int jj = cj; jj + 1 < m; jj++)
        g[ci][jj].push_back({ci, jj + 1}), res[ci][jj + 1]++;
      for (int jj = cj; jj - 1 >= 0; jj--)
        g[ci][jj].push_back({ci, jj - 1}), res[ci][jj - 1]++;
    }
  }
  priority_queue<pair<int, int>, vector<pair<int, int> >, comp> pq;
  for (int v = n * m; v; v--) {
    pair<int, int> p;
    if (!u[v]) {
      u[v] = 1;
      p = pq.top();
      pq.pop();
      vis[p.first][p.second] = 1;
      a[p.first][p.second] = v;
      pos[v] = p;
    } else
      p = pos[v];
    for (pair<int, int> q : g[p.first][p.second]) {
      res[q.first][q.second]--;
      if (!res[q.first][q.second] && !vis[q.first][q.second]) {
        pq.push(q);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) printf("%d ", a[i][j]);
    printf("\n");
  }
}