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;
int N, M;
int A[300][300], B[300][300];
int R[300], C[300];
bool usd[250 * 250 + 5];
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) scanf("%d", &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]);
  }
  for (int j = 0; j < M; j++) {
    for (int i = 0; i < N; i++) C[j] = max(C[j], A[i][j]);
  }
  sort(R, R + N);
  sort(C, C + M);
  reverse(R, R + N);
  reverse(C, C + M);
  int ri = 0, ci = 0;
  while (ri < N || ci < M) {
    if (R[ri] == C[ci]) {
      B[ri][ci] = R[ri];
      usd[R[ri]] = true;
      ri++;
      ci++;
    } else if (R[ri] < C[ci]) {
      B[ri - 1][ci] = C[ci];
      usd[C[ci]] = true;
      ci++;
    } else {
      B[ri][ci - 1] = R[ri];
      usd[R[ri]] = true;
      ri++;
    }
  }
  int id = 1;
  ri--, ci--;
  while (ri >= 0 && ci >= 0) {
    if (R[ri] <= C[ci]) {
      for (int k = 0; k <= ci; k++)
        if (B[ri][k] == 0) {
          while (usd[id]) id++;
          B[ri][k] = id;
          usd[id] = true;
        }
      ri--;
    } else {
      for (int k = 0; k <= ri; k++)
        if (B[k][ci] == 0) {
          while (usd[id]) id++;
          B[k][ci] = id;
          usd[id] = true;
        }
      ci--;
    }
  }
  bool check = true;
  for (int i = 0; i < N; i++) {
    int mR = 0;
    for (int j = 0; j < M; j++) mR = max(mR, B[i][j]);
    if (mR != R[i]) check = false;
  }
  for (int j = 0; j < M; j++) {
    int mC = 0;
    for (int i = 0; i < N; i++) mC = max(mC, B[i][j]);
    if (mC != C[j]) check = false;
  }
  if (check) {
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < M; j++)
        printf("%d%c", B[i][j], j + 1 == M ? '\n' : ' ');
    }
  } else
    puts("-1");
}