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")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,mmx,avx")
using namespace std;
const int MOD = 1000000007;
const int UNDEF = -1;
const int INF = 1 << 30;
template <typename T>
inline bool chkmax(T& aa, T bb) {
  return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T& aa, T bb) {
  return aa > bb ? aa = bb, true : false;
}
int rint();
char rch();
long long rlong();
void fail() {
  printf("-1\n");
  exit(0);
}
int ur = 0, uc = 0;
set<pair<int, int> > good;
set<pair<int, int> > bad;
int rlim, clim;
vector<vector<int> > ans;
int dr[] = {1, 0, -1, 0};
int dc[] = {0, 1, 0, -1};
bool isgood(int r, int c) {
  int gr = 0, gc = 0;
  for (int nr = r - 1; nr <= r + 1; nr += 2) {
    if (1 <= nr && nr <= rlim && ans[nr][c]) gr++;
  }
  for (int nc = c - 1; nc <= c + 1; nc += 2) {
    if (1 <= nc && nc <= clim && ans[r][nc]) gc++;
  }
  return gr && gc;
}
void paint(int r, int c, int val) {
  ans[r][c] = val;
  good.erase(make_pair(r, c));
  bad.erase(make_pair(r, c));
  for (int k = 0; k < 4; k++) {
    int nr = r + dr[k], nc = c + dc[k];
    auto it = bad.find(make_pair(nr, nc));
    if (it != bad.end() && isgood(nr, nc)) {
      bad.erase(it);
      good.insert(make_pair(nr, nc));
    }
  }
}
void ins(int r, int c) {
  if (isgood(r, c))
    good.insert(make_pair(r, c));
  else
    bad.insert(make_pair(r, c));
}
void incr() {
  ++ur;
  for (int c = 1; c <= uc; c++) ins(ur, c);
}
void incc() {
  ++uc;
  for (int r = 1; r <= ur; r++) ins(r, uc);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> rlim >> clim;
  vector<vector<int> > inp(rlim + 1);
  ans.resize(rlim + 1);
  vector<int> rmax(rlim + 1), cmax(clim + 1);
  for (int r = 1; r <= rlim; r++) {
    inp[r].resize(clim + 1);
    ans[r].resize(clim + 1);
    for (int c = 1; c <= clim; c++) {
      int e;
      cin >> e;
      inp[r][c] = e;
      chkmax(rmax[r], e);
      chkmax(cmax[c], e);
    }
  }
  vector<int> hi2r(rlim * clim + 1), hi2c(rlim * clim + 1);
  for (int r = 1; r <= rlim; r++) {
    hi2r[rmax[r]] = r;
  }
  for (int c = 1; c <= clim; c++) {
    hi2c[cmax[c]] = c;
  }
  for (int val = rlim * clim; val >= 1; val--) {
    int ro = hi2r[val], co = hi2c[val];
    if ((!ro) && (!co)) {
      auto it = good.begin();
      if (it == good.end()) fail();
      int rn = it->first, cn = it->second;
      paint(rn, cn, val);
    } else {
      if (ro && co) {
        incr();
        incc();
      } else if (ro) {
        incr();
      } else if (co) {
        incc();
      } else {
        ;
      }
      paint(ur, uc, val);
    }
  }
  for (int r = 1; r <= rlim; r++) {
    for (int c = 1; c <= clim; c++) {
      printf("%d ", ans[r][c]);
    }
    printf("\n");
  }
}
static char stdinBuffer[1024];
static char* stdinDataEnd = stdinBuffer + sizeof(stdinBuffer);
static const char* stdinPos = stdinDataEnd;
void readAhead(size_t amount) {
  size_t remaining = stdinDataEnd - stdinPos;
  if (remaining < amount) {
    memmove(stdinBuffer, stdinPos, remaining);
    size_t sz = fread(stdinBuffer + remaining, 1,
                      sizeof(stdinBuffer) - remaining, stdin);
    stdinPos = stdinBuffer;
    stdinDataEnd = stdinBuffer + remaining + sz;
    if (stdinDataEnd != stdinBuffer + sizeof(stdinBuffer)) *stdinDataEnd = 0;
  }
}
int rint() {
  readAhead(16);
  int x = 0;
  bool neg = false;
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  if (*stdinPos == '-') {
    ++stdinPos;
    neg = true;
  }
  while (*stdinPos >= '0' && *stdinPos <= '9') {
    x *= 10;
    x += *stdinPos - '0';
    ++stdinPos;
  }
  return neg ? -x : x;
}
char rch() {
  readAhead(16);
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  char ans = *stdinPos;
  ++stdinPos;
  return ans;
}
long long rlong() {
  readAhead(32);
  long long x = 0;
  bool neg = false;
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  if (*stdinPos == '-') {
    ++stdinPos;
    neg = true;
  }
  while (*stdinPos >= '0' && *stdinPos <= '9') {
    x *= 10;
    x += *stdinPos - '0';
    ++stdinPos;
  }
  return neg ? -x : x;
}