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;
long long mod(long long a) {
  long long c = a % (long long)(1E9 + 7);
  if (c >= 0) return c;
  return c + (long long)(1E9 + 7);
}
long long inf = (long long)1E16;
long long max(long long a, long long b) {
  if (a < b)
    return b;
  else
    return a;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
void Max(long long &pos, long long val) { pos = max(pos, val); }
void Min(long long &pos, long long val) { pos = min(pos, val); }
void Add(long long &pos, long long val) { pos = mod(pos + val); }
long double EPS = 1E-9;
template <class T, class S>
inline bool chmax(T &a, S b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class S>
inline bool chmin(T &a, S b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
void mod_print(long long k) {
  long long P = 50000;
  for (long long y = 1; y <= P; y++) {
    long long x = mod(y * k);
    if (abs(x) <= P || x + P >= (long long)(1E9 + 7)) {
      if (x + P >= (long long)(1E9 + 7)) {
        x -= (long long)(1E9 + 7);
      }
      cout << x << "/" << y << "\n";
      return;
    }
  }
  cout << "nun"
       << "\n";
}
long long n, m;
bool valid(long long i, long long j) {
  if (0 <= i && i <= n - 1 && 0 <= j && j <= m - 1) return true;
  return false;
}
long long yo[255][255];
long long ta[255][255];
long long ans[255][255];
long long dx[4] = {1, 0, -1, 0};
long long dy[4] = {0, 1, 0, -1};
bool no[255][255];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  long long z[n][m];
  for (long long i = 0; i <= n - 1; i++)
    for (long long j = 0; j <= m - 1; j++) cin >> z[i][j];
  vector<long long> a(n), b(m);
  for (long long i = 0; i <= n - 1; i++) a[i] = 0;
  for (long long i = 0; i <= m - 1; i++) b[i] = 0;
  for (long long i = 0; i <= n - 1; i++) {
    for (long long j = 0; j <= m - 1; j++) chmax(a[i], z[i][j]);
    for (long long j = 0; j <= m - 1; j++) chmax(b[j], z[i][j]);
  }
  long long c[n * m + 1];
  memset(c, 0, sizeof(c));
  for (long long i = 0; i <= n - 1; i++) c[a[i]]++;
  for (long long i = 0; i <= m - 1; i++) c[b[i]] += 2;
  vector<long long> d, e, f;
  for (long long i = 1; i <= n * m; i++) {
    if (c[i] == 0) continue;
    if (c[i] == 3)
      d.push_back(i);
    else if (c[i] == 1)
      e.push_back(i);
    else if (c[i] == 2)
      f.push_back(i);
  }
  sort(d.begin(), d.end());
  sort(e.begin(), e.end());
  sort(f.begin(), f.end());
  reverse(e.begin(), e.end());
  reverse(f.begin(), f.end());
  memset(ans, -1, sizeof(ans));
  long long ds = d.size();
  pair<long long, long long> mp[n * m + 1];
  for (long long i = 1; i <= n * m; i++) mp[i] = {-1, -1};
  for (long long i = 0; i <= ds - 1; i++) {
    ans[i][i] = d[i];
    mp[d[i]] = {i, i};
  }
  for (long long i = ds; i <= n - 1; i++) {
    ans[i][ds - 1] = e[i - ds];
    mp[e[i - ds]] = {i, ds - 1};
  }
  for (long long i = ds; i <= m - 1; i++) {
    ans[ds - 1][i] = f[i - ds];
    mp[f[i - ds]] = {ds - 1, i};
  }
  memset(no, false, sizeof(no));
  for (long long i = 1; i <= n * m; i++) {
    if (mp[i].first != -1) {
      no[mp[i].first][mp[i].second] = true;
    }
  }
  queue<pair<long long, long long> > q;
  for (long long i = n * m; i >= 1; i--) {
    if (mp[i].first == -1) {
      pair<long long, long long> t;
      while (1) {
        t = q.front();
        q.pop();
        if (ans[t.first][t.second] == -1) break;
      }
      ans[t.first][t.second] = i;
      long long x = t.first;
      long long y = t.second;
      for (long long j = -1; j <= 1; j += 2) {
        long long nx = x + j;
        if (valid(nx, y)) {
          if (no[nx][y]) continue;
          ta[nx][y]++;
          if (yo[nx][y]) q.push({nx, y});
        }
      }
      for (long long j = -1; j <= 1; j += 2) {
        long long ny = y + j;
        if (valid(x, ny)) {
          if (no[x][ny]) continue;
          yo[x][ny]++;
          if (ta[x][ny]) q.push({x, ny});
        }
      }
    } else {
      long long x = mp[i].first;
      long long y = mp[i].second;
      for (long long j = -1; j <= 1; j += 2) {
        long long nx = x + j;
        if (valid(nx, y)) {
          if (no[nx][y]) continue;
          ta[nx][y]++;
          if (yo[nx][y]) q.push({nx, y});
        }
      }
      for (long long j = -1; j <= 1; j += 2) {
        long long ny = y + j;
        if (valid(x, ny)) {
          if (no[x][ny]) continue;
          yo[x][ny]++;
          if (ta[x][ny]) q.push({x, ny});
        }
      }
    }
  }
  for (long long i = 0; i <= n - 1; i++) {
    for (long long j = 0; j <= m - 1; j++) cout << ans[i][j] << " ";
    cout << "\n";
  }
  return 0;
}