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>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
  for (bool sp = 0; begin != end; ++begin) {
    if (sp)
      putchar(' ');
    else
      sp = true;
    printf(fmt, *begin);
  }
  putchar('\n');
}
int N, M;
int A[255][255];
int B[255][255];
int R[255], C[255];
bool use[100011];
bool in[255][255];
void MAIN() {
  scanf("%d%d", &N, &M);
  for (int i = 0, i_len = (N); i < i_len; ++i)
    for (int j = 0, j_len = (M); j < j_len; ++j) scanf("%d", &A[i][j]);
  for (int i = 0, i_len = (N); i < i_len; ++i)
    for (int j = 0, j_len = (M); j < j_len; ++j) {
      amax(R[i], A[i][j]);
      amax(C[j], A[i][j]);
    }
  sort(R, R + N);
  sort(C, C + M);
  int ri = N - 1, ci = M - 1;
  int pos_r = 0, pos_c = 0;
  while (1) {
    if (ri < 0 && ci < 0) break;
    int r_value = (ri < 0 ? -1 : R[ri]);
    int c_value = (ci < 0 ? -1 : C[ci]);
    if (r_value == c_value) {
      B[pos_r][pos_c] = r_value;
      pos_r++;
      pos_c++;
      ri--;
      ci--;
    } else if (r_value < c_value) {
      B[pos_r - 1][pos_c] = c_value;
      pos_c++;
      ci--;
    } else {
      B[pos_r][pos_c - 1] = r_value;
      pos_r++;
      ri--;
    }
  }
  for (int i = 0, i_len = (N); i < i_len; ++i)
    for (int j = 0, j_len = (M); j < j_len; ++j) use[B[i][j]] = true;
  using Tuple = tuple<int, int, int>;
  priority_queue<Tuple> Q;
  auto update = [&](int i, int j) {
    if (B[i][j] == 0 && !in[i][j]) {
      if (0 < i && j + 1 < M && B[i - 1][j] && B[i][j + 1]) {
        in[i][j] = true;
        Q.emplace(min(B[i - 1][j], B[i][j + 1]), i, j);
      }
      if (i + 1 < N && 0 < j && B[i + 1][j] && B[i][j - 1]) {
        in[i][j] = true;
        Q.emplace(min(B[i + 1][j], B[i][j - 1]), i, j);
      }
    }
  };
  for (int i = 0, i_len = (N); i < i_len; ++i)
    for (int j = 0, j_len = (M); j < j_len; ++j) update(i, j);
  int idx = N * M;
  while (!Q.empty()) {
    int v, i, j;
    tie(v, i, j) = Q.top();
    Q.pop();
    while (use[idx]) idx--;
    B[i][j] = idx;
    use[idx] = true;
    if (i) update(i - 1, j);
    if (j) update(i, j - 1);
    if (i + 1 < N) update(i + 1, j);
    if (j + 1 < M) update(i, j + 1);
  }
  for (int i = 0, i_len = (N); i < i_len; ++i) rprintf("%d", B[i], B[i] + M);
}
int main() {
  int TC = 1;
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}