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 = 301;
int n, m, a[N][N], r[N], c[N], vr[N * N], vc[N * N];
int ans[N][N], lim[N][N];
struct node {
  int x, y, v;
} s[N * N];
bool cmp(node a, node b) {
  if (a.v != b.v) return a.v < b.v;
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      r[i] = max(r[i], a[i][j]), c[j] = max(c[j], a[i][j]);
  for (int i = 1; i <= n; i++) vr[r[i]] = 1;
  for (int i = 1; i <= m; i++) vc[c[i]] = 1;
  int nr = 0, nc = 0;
  for (int i = n * m; i >= 1; i--)
    if (vr[i] || vc[i]) {
      if (vr[i]) nr++;
      if (vc[i]) nc++;
      ans[nr][nc] = i;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) lim[i][j] = 1e9;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (ans[i][j]) {
        for (int k = 1; k <= m; k++)
          if (!ans[i][k]) lim[i][k] = min(lim[i][k], ans[i][j]);
        for (int k = 1; k <= n; k++)
          if (!ans[k][j]) lim[k][j] = min(lim[k][j], ans[i][j]);
      }
  int tot = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (!ans[i][j]) {
        node tmp;
        tmp.x = i, tmp.y = j, tmp.v = lim[i][j];
        s[++tot] = tmp;
      }
  sort(s + 1, s + tot + 1, cmp);
  int now = 1;
  int ok = 1;
  for (int i = 1; i <= tot; i++) {
    while (vr[now] || vc[now]) now++;
    if (s[i].v < now) ok = 0;
    ans[s[i].x][s[i].y] = now;
    now++;
  }
  if (!ok)
    puts("-1");
  else {
    for (int i = 1; i <= n; i++, puts(""))
      for (int j = 1; j <= m; j++, putchar(' ')) printf("%d", ans[i][j]);
  }
}