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 N = 255;
int n, m;
int a[N][N], b[N][N];
int x[N], y[N], lx[N], rx[N], ly[N], ry[N];
bool visx[N], visy[N];
pair<int, int> p[N * N];
int fx[4] = {0, 0, -1, 1};
int fy[4] = {-1, 1, 0, 0};
void add(int x1, int y1, int val) {
  b[x1][y1] = val;
  visx[x1] = true;
  visy[y1] = true;
  lx[x1] = min(lx[x1], y1);
  rx[x1] = max(rx[x1], y1);
  ly[y1] = min(ly[y1], x1);
  ry[y1] = max(ry[y1], x1);
}
bool check(int x, int y) { return x >= 1 && x <= n && y >= 1 && y <= m; }
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      cin >> a[i][j];
      x[i] = max(x[i], a[i][j]);
      y[j] = max(y[j], a[i][j]);
    }
  memset(lx, 0x3f, sizeof(lx));
  memset(ly, 0x3f, sizeof(ly));
  memset(rx, -0x3f, sizeof(rx));
  memset(ry, -0x3f, sizeof(ry));
  for (int i = 1; i <= n; ++i) p[x[i]].first = 1;
  for (int i = 1; i <= m; ++i) p[y[i]].second = 1;
  for (int i = n * m; i >= 1; --i) {
    if (p[i].first && p[i].second) {
      for (int j = 1; j <= n; ++j)
        if (!visx[j]) {
          for (int k = 1; k <= m; ++k)
            if (!visy[k]) {
              add(j, k, i);
              break;
            }
          break;
        }
    } else if (p[i].first) {
      for (int j = 1; j <= m; ++j)
        if (check(ly[j] - 1, j) && !visx[ly[j] - 1]) {
          add(ly[j] - 1, j, i);
          break;
        } else if (check(ry[j] + 1, j) && !visx[ry[j] + 1]) {
          add(ry[j] + 1, j, i);
          break;
        }
    } else if (p[i].second) {
      for (int j = 1; j <= n; ++j)
        if (check(j, lx[j] - 1) && !visy[lx[j] - 1]) {
          add(j, lx[j] - 1, i);
          break;
        } else if (check(j, rx[j] + 1) && !visy[rx[j] + 1]) {
          add(j, rx[j] + 1, i);
          break;
        }
    } else {
      for (int j = 1; j <= m; ++j)
        if (check(ly[j] - 1, j) && visx[ly[j] - 1]) {
          int x = ly[j] - 1;
          if (lx[x] - 1 != j && rx[x] + 1 != j) continue;
          add(ly[j] - 1, j, i);
          break;
        } else if (check(ry[j] + 1, j) && visx[ry[j] + 1]) {
          int x = ry[j] + 1;
          if (lx[x] - 1 != j && rx[x] + 1 != j) continue;
          add(ry[j] + 1, j, i);
          break;
        }
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (!b[i][j]) {
        printf("-1");
        return 0;
      }
  for (int j = 1; j <= n; ++j) {
    for (int k = 1; k <= m; ++k) {
      printf("%d ", b[j][k]);
    }
    printf("\n");
  }
}