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 int MX = 255;
int n, m, a[MX][MX], res[MX][MX], vis[MX * MX];
vector<int> col, fil;
set<int> st;
set<pair<int, pair<int, int>>> pq;
int mx[] = {1, -1, 0, 0};
int my[] = {0, 0, 1, -1};
int valid(int i, int j) {
  if (i < 0 || n <= i || j < 0 || m <= j || !res[i][j]) return -1;
  return res[i][j];
}
void candi(int i, int j) {
  if (i < 0 || n <= i || j < 0 || m <= j || res[i][j]) return;
  int cn = 0, mn = MX * MX;
  for (int l = 0; l < 4; l++) {
    int t = valid(i + mx[l], j + my[l]);
    if (t != -1) mn = min(mn, t), cn++;
  }
  if (cn == 2) pq.emplace(mn, pair<int, int>(i, j));
}
void solve() {
  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 * m; i++) st.insert(i + 1);
  for (int i = 0; i < n; i++) {
    int mx = 0;
    for (int j = 0; j < m; j++) mx = max(mx, a[i][j]);
    col.push_back(mx);
    st.erase(mx);
  }
  for (int j = 0; j < m; j++) {
    int mx = 0;
    for (int i = 0; i < n; i++) mx = max(mx, a[i][j]);
    fil.push_back(mx);
    st.erase(mx);
  }
  sort(col.begin(), col.end(), greater<int>());
  sort(fil.begin(), fil.end(), greater<int>());
  int x = 0, y = 0, ax = 0, ay = 0;
  for (int i = 0, j = 0; i < col.size() || j < fil.size();) {
    if (i < col.size() && j < fil.size() && col[i] == fil[j]) {
      res[x][y] = col[i];
      ax = x, ay = y;
      x++, y++;
      i++, j++;
    } else if (i < col.size() && (j == fil.size() || col[i] > fil[j])) {
      res[x][ay] = col[i];
      x++;
      i++;
    } else {
      res[ax][y] = fil[j];
      y++;
      j++;
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) candi(i, j);
  while (st.size()) {
    int k = *st.rbegin(), i, j;
    auto it = *pq.rbegin();
    tie(i, j) = it.second;
    res[i][j] = k;
    for (int l = 0; l < 4; l++) candi(i + mx[l], j + my[l]);
    st.erase(k);
    pq.erase(it);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << res[i][j] << " ";
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}