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("Ofast")
using namespace std;
inline int my_getchar() {
  static char buf[1048576];
  static int s = 1048576;
  static int e = 1048576;
  if (s == e && e == 1048576) {
    e = fread(buf, 1, 1048576, stdin);
    s = 0;
  }
  if (s == e) {
    return EOF;
  }
  return buf[s++];
}
inline void rd(int &x) {
  int k;
  int m = 0;
  x = 0;
  for (;;) {
    k = my_getchar();
    if (k == '-') {
      m = 1;
      break;
    }
    if ('0' <= k && k <= '9') {
      x = k - '0';
      break;
    }
  }
  for (;;) {
    k = my_getchar();
    if (k < '0' || k > '9') {
      break;
    }
    x = x * 10 + k - '0';
  }
  if (m) {
    x = -x;
  }
}
struct MY_WRITER {
  char buf[1048576];
  int s;
  int e;
  MY_WRITER() {
    s = 0;
    e = 1048576;
  }
  ~MY_WRITER() {
    if (s) {
      fwrite(buf, 1, s, stdout);
    }
  }
};
MY_WRITER MY_WRITER_VAR;
void my_putchar(int a) {
  if (MY_WRITER_VAR.s == MY_WRITER_VAR.e) {
    fwrite(MY_WRITER_VAR.buf, 1, MY_WRITER_VAR.s, stdout);
    MY_WRITER_VAR.s = 0;
  }
  MY_WRITER_VAR.buf[MY_WRITER_VAR.s++] = a;
}
inline void wt_L(char a) { my_putchar(a); }
inline void wt_L(int x) {
  int s = 0;
  int m = 0;
  char f[10];
  if (x < 0) {
    m = 1;
    x = -x;
  }
  while (x) {
    f[s++] = x % 10;
    x /= 10;
  }
  if (!s) {
    f[s++] = 0;
  }
  if (m) {
    my_putchar('-');
  }
  while (s--) {
    my_putchar(f[s] + '0');
  }
}
template <class S>
inline void arrInsert(const int k, int &sz, S a[], const S aval) {
  int i;
  sz++;
  for (i = sz - 1; i > k; i--) {
    a[i] = a[i - 1];
  }
  a[k] = aval;
}
template <class S, class T>
inline void arrInsert(const int k, int &sz, S a[], const S aval, T b[],
                      const T bval) {
  int i;
  sz++;
  for (i = sz - 1; i > k; i--) {
    a[i] = a[i - 1];
  }
  for (i = sz - 1; i > k; i--) {
    b[i] = b[i - 1];
  }
  a[k] = aval;
  b[k] = bval;
}
template <class S, class T, class U>
inline void arrInsert(const int k, int &sz, S a[], const S aval, T b[],
                      const T bval, U c[], const U cval) {
  int i;
  sz++;
  for (i = sz - 1; i > k; i--) {
    a[i] = a[i - 1];
  }
  for (i = sz - 1; i > k; i--) {
    b[i] = b[i - 1];
  }
  for (i = sz - 1; i > k; i--) {
    c[i] = c[i - 1];
  }
  a[k] = aval;
  b[k] = bval;
  c[k] = cval;
}
template <class S, class T, class U, class V>
inline void arrInsert(const int k, int &sz, S a[], const S aval, T b[],
                      const T bval, U c[], const U cval, V d[], const V dval) {
  int i;
  sz++;
  for (i = sz - 1; i > k; i--) {
    a[i] = a[i - 1];
  }
  for (i = sz - 1; i > k; i--) {
    b[i] = b[i - 1];
  }
  for (i = sz - 1; i > k; i--) {
    c[i] = c[i - 1];
  }
  for (i = sz - 1; i > k; i--) {
    d[i] = d[i - 1];
  }
  a[k] = aval;
  b[k] = bval;
  c[k] = cval;
  d[k] = dval;
}
template <class S, class T>
inline S chmax(S &a, T b) {
  if (a < b) {
    a = b;
  }
  return a;
}
int N;
int M;
int A[250][250];
int rm[250];
int cm[250];
int ind[100000];
int r[100000];
int c[100000];
int qs;
int qr[100000];
int qc[100000];
void doit(int rr, int cc, int v) {
  int di;
  int dj;
  int i;
  int j;
  A[rr][cc] = v;
  for (di = (-2); di < (3); di++) {
    for (dj = (-2); dj < (3); dj++) {
      i = rr + di;
      j = cc + dj;
      if (i < 0 || i >= N || j < 0 || j >= M || A[i][j] != -1) {
        continue;
      }
      if (i - 1 >= 0 && j + 1 < M && A[i - 1][j] >= 0 && A[i][j + 1] >= 0) {
        A[i][j] = -2;
        arrInsert(qs, qs, qr, i, qc, j);
        continue;
      }
      if (i + 1 < N && j - 1 >= 0 && A[i + 1][j] >= 0 && A[i][j - 1] >= 0) {
        A[i][j] = -2;
        arrInsert(qs, qs, qr, i, qc, j);
        continue;
      }
    }
  }
}
int main() {
  int i;
  int rr = 0;
  int cc = 0;
  rd(N);
  rd(M);
  {
    int cTE1_r3A;
    int RZTsC2BF;
    for (cTE1_r3A = (0); cTE1_r3A < (N); cTE1_r3A++) {
      for (RZTsC2BF = (0); RZTsC2BF < (M); RZTsC2BF++) {
        rd(A[cTE1_r3A][RZTsC2BF]);
        A[cTE1_r3A][RZTsC2BF] += (-1);
      }
    }
  }
  for (i = (0); i < (N); i++) {
    int j;
    for (j = (0); j < (M); j++) {
      chmax(rm[i], A[i][j]);
    }
  }
  for (i = (0); i < (N); i++) {
    int j;
    for (j = (0); j < (M); j++) {
      chmax(cm[j], A[i][j]);
    }
  }
  for (i = (0); i < (N); i++) {
    r[rm[i]] = 1;
  }
  for (i = (0); i < (M); i++) {
    c[cm[i]] = 1;
  }
  for (i = (0); i < (N); i++) {
    int j;
    for (j = (0); j < (M); j++) {
      A[i][j] = -1;
    }
  }
  for (i = (N * M) - 1; i >= (0); i--) {
    if (r[i] && c[i]) {
      doit(rr++, cc++, i);
      continue;
    }
    if (r[i]) {
      doit(rr++, cc - 1, i);
      continue;
    }
    if (c[i]) {
      doit(rr - 1, cc++, i);
      continue;
    }
    qs--;
    doit(qr[qs], qc[qs], i);
  }
  {
    int gEg5UqEA;
    int qSsg05KM;
    for (gEg5UqEA = (0); gEg5UqEA < (N); gEg5UqEA++) {
      if (M == 0) {
        wt_L('\n');
      } else {
        for (qSsg05KM = (0); qSsg05KM < (M - 1); qSsg05KM++) {
          wt_L(A[gEg5UqEA][qSsg05KM] + 1);
          wt_L(' ');
        }
        wt_L(A[gEg5UqEA][qSsg05KM] + 1);
        wt_L('\n');
      }
    }
  }
  return 0;
}