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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 1e9 + 7;
double eps = 1e-12;
long long mpow(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2);
  t1 *= t1;
  t1 %= MOD;
  if (b % 2) t1 *= a;
  t1 %= MOD;
  return t1;
}
long long mpow(long long a, long long b, long long p) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2, p);
  t1 *= t1;
  t1 %= p;
  if (b % 2) t1 *= a;
  t1 %= p;
  return t1;
}
long long modinverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
long long range(long long l, long long r) { return l + mt() % (r - l + 1); }
long long rev(long long v) { return mpow(v, MOD - 2); }
void solve() {
  long long n, m;
  cin >> n >> m;
  long long a[n][m];
  long long r[n], c[m];
  for (long long i = 0; i < n; i++) r[i] = 0;
  for (long long i = 0; i < m; i++) c[i] = 0;
  long long ans[n][m], nb[n][m];
  long long val[n * m + 1];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cin >> a[i][j];
      nb[i][j] = ans[i][j] = 0;
      val[i * m + j + 1] = 0;
      r[i] = max(r[i], a[i][j]);
      c[j] = max(c[j], a[i][j]);
    }
  }
  sort(r, r + n);
  sort(c, c + m);
  long long ri = n - 1, ci = m - 1;
  long long cr = -1, cc = -1;
  pair<long long, long long> pos[n * m + 1];
  while (ri != -1 || ci != -1) {
    if (ri == -1) {
      cc++;
      ans[cr][cc] = c[ci];
      ci--;
    } else if (ci == -1 || (r[ri] > c[ci])) {
      cr++;
      ans[cr][cc] = r[ri];
      ri--;
    } else if (c[ci] > r[ri]) {
      cc++;
      ans[cr][cc] = c[ci];
      ci--;
    } else {
      cr++;
      cc++;
      ans[cr][cc] = r[ri];
      ri--;
      ci--;
    }
    pos[ans[cr][cc]] = make_pair(cr, cc);
    val[ans[cr][cc]] = 1;
  }
  vector<pair<long long, long long> > diff;
  diff.push_back(make_pair(-1, 0));
  diff.push_back(make_pair(1, 0));
  diff.push_back(make_pair(0, -1));
  diff.push_back(make_pair(0, 1));
  vector<pair<long long, long long> > vec;
  for (long long k = n * m; k >= 1; k--) {
    if (!val[k]) {
      assert(((long long)(vec).size()));
      pair<long long, long long> lst = vec.back();
      vec.pop_back();
      ans[lst.first][lst.second] = k;
      for (auto it : diff) {
        long long nx = (lst.first + it.first), ny = (lst.second + it.second);
        if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
          nb[nx][ny]++;
          if (nb[nx][ny] == 2 && !ans[nx][ny]) vec.push_back(make_pair(nx, ny));
        }
      }
    } else {
      pair<long long, long long> lst = pos[k];
      for (auto it : diff) {
        long long nx = (lst.first + it.first), ny = (lst.second + it.second);
        if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
          nb[nx][ny]++;
          if (nb[nx][ny] == 2 && !ans[nx][ny]) vec.push_back(make_pair(nx, ny));
        }
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) cout << ans[i][j] << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
