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 \}) 
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937 rng(steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
long long n, m, k, q, l, r, x, y, z;
long long a[1000005];
long long b[1000005];
long long c[1000005];
string second, t;
long long ans = 0;
long long mat[255][255];
bool vis[255][255];
long long mr[255], mc[255];
long long cnt[255][255];
long long omat[255][255];
const long long dr[4] = {0, 0, -1, 1};
const long long dc[4] = {1, -1, 0, 0};
bool valid(long long r, long long c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
void solve(int tt) {
  cin >> n >> m;
  for (long long i = 0; i < (n); ++i)
    for (int ele = 0; ele < m; ele++) cin >> mat[i][ele];
  ;
  memset(vis, 0, sizeof(vis));
  memset(mr, 0, sizeof(mr));
  memset(mc, 0, sizeof(mc));
  for (long long i = 0; i < (n); ++i) {
    for (long long j = 0; j < (m); ++j) {
      --mat[i][j];
      mr[i] = max(mr[i], mat[i][j]);
      mc[j] = max(mc[j], mat[i][j]);
    }
  }
  memset(cnt, 0, sizeof(cnt));
  pair<long long, long long> pos = make_pair(-1, -1);
  for (long long i = 0; i < (n); ++i)
    for (long long j = 0; j < (m); ++j) omat[i][j] = -1;
  set<pair<long long, long long> > safe;
  for (long long i = (n * m - 1); i >= 0; --i) {
    bool rr = 0, rc = 0;
    for (long long j = 0; j < (n); ++j) {
      if (mr[j] == i) {
        rr = 1;
      }
    }
    for (long long j = 0; j < (m); ++j) {
      if (mc[j] == i) {
        rc = 1;
      }
    }
    if (rr) ++pos.first;
    if (rc) ++pos.second;
    pair<long long, long long> loc = pos;
    if (!rr && !rc) {
      if (!safe.size()) {
        cout << -1 << '\n';
        return;
      }
      loc = *safe.begin();
      safe.erase(safe.begin());
    }
    omat[loc.first][loc.second] = i + 1;
    for (long long d = 0; d < (4); ++d) {
      long long ni = loc.first + dr[d], nj = loc.second + dc[d];
      if (valid(ni, nj)) {
        ++cnt[ni][nj];
        if (cnt[ni][nj] == 2 && omat[ni][nj] == -1) {
          safe.insert(make_pair(ni, nj));
        }
      }
    }
  }
  for (long long i = 0; i < (n); ++i) {
    for (int ele = 0; ele < (m); ele++) {
      if (ele) cout << " ";
      cout << omat[i][ele];
    }
    cout << '\n';
  };
}
int main() {
  { ios_base::sync_with_stdio(false); }
  {
    cin.tie(NULL);
    cout.tie(NULL);
  }
  int tc = 1;
  for (int t = 0; t < tc; t++) solve(t);
}
