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;
const long long maxn = 1e3 + 100;
const long long inf = 0x3f3f3f3f;
const long long iinf = 1 << 30;
const long long linf = 2e18;
const long long mod = 998244353;
const double eps = 1e-7;
template <class T = long long>
T chmin(T &a, T b) {
  return a = min(a, b);
}
template <class T = long long>
T chmax(T &a, T b) {
  return a = max(a, b);
}
template <class T = long long>
inline void red(T &x) {
  x -= mod, x += x >> 31 & mod;
}
template <class T = long long>
T read() {
  T f = 1, a = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    a = (a << 3) + (a << 1) + ch - '0';
    ch = getchar();
  }
  return a * f;
}
long long n, m;
long long a[maxn][maxn], c[maxn][maxn], r[maxn][maxn];
long long typ[maxn * maxn], ans[maxn][maxn];
long long dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
priority_queue<pair<long long, pair<long long, long long> > > q;
signed main() {
  n = read(), m = read();
  for (long long i = (1); i <= (n); ++i)
    for (long long j = (1); j <= (m); ++j) a[i][j] = read();
  for (long long i = (1); i <= (n); ++i)
    for (long long j = (1); j <= (m); ++j)
      c[i][j] = max(c[i][j - 1], a[i][j]), r[i][j] = max(r[i - 1][j], a[i][j]);
  for (long long i = (1); i <= (n); ++i)
    for (long long j = (1); j <= (m); ++j) {
      if (a[i][j] == c[i][m]) typ[a[i][j]] |= 1;
      if (a[i][j] == r[n][j]) typ[a[i][j]] |= 2;
    }
  long long cx = 0, cy = 0;
  for (long long i = (n * m); i >= (1); --i) {
    if (!typ[i]) continue;
    if (typ[i] & 1) cx++;
    if (typ[i] & 2) cy++;
    ans[cx][cy] = i;
  }
  for (long long i = (1); i <= (n); ++i)
    for (long long j = (1); j <= (m); ++j) {
      if (ans[i][j]) continue;
      if (ans[i - 1][j] && ans[i][j + 1])
        q.push(make_pair(min(ans[i - 1][j], ans[i][j + 1]), make_pair(i, j)));
      if (ans[i + 1][j] && ans[i][j - 1])
        q.push(make_pair(min(ans[i + 1][j], ans[i][j - 1]), make_pair(i, j)));
    }
  long long tot = n * m;
  while (!q.empty()) {
    long long nx = q.top().second.first, ny = q.top().second.second,
              val = q.top().first;
    q.pop();
    while (tot > 1 && typ[tot]) --tot;
    assert(tot), assert(tot < val);
    ans[nx][ny] = tot;
    tot--;
    for (long long k = (0); k <= (3); ++k) {
      long long tx = nx + dx[k], ty = ny + dy[k];
      if (tx < 1 || ty < 1 || tx > n || ty > m) continue;
      if (ans[tx][ty]) continue;
      if (ans[tx - 1][ty] && ans[tx][ty + 1])
        q.push(make_pair(min(ans[tx - 1][ty], ans[tx][ty + 1]),
                         make_pair(tx, ty)));
      if (ans[tx + 1][ty] && ans[tx][ty - 1])
        q.push(make_pair(min(ans[tx + 1][ty], ans[tx][ty - 1]),
                         make_pair(tx, ty)));
    }
  }
  for (long long i = (1); i <= (n); ++i)
    for (long long j = (1); j <= (m); ++j)
      printf("%lld%c", ans[i][j], " \n"[j == m]);
  return 0;
}
