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;
class Dumae {
 public:
  int n, m, K, be[301000], ed[301000], Q[301000], head, tail, DegE[301000],
      DegF[301000], Res[301000], inp[301000];
  int RRes[301000];
  vector<int> E[301000], F[301000], G[301000];
  void Do(vector<pair<int, int> > Ed, vector<pair<int, int> > Ran) {
    int i, a, b;
    n = Ran.size();
    m = Ed.size();
    for (i = 1; i <= n; i++) {
      be[i] = Ran[i - 1].first, ed[i] = Ran[i - 1].second;
    }
    for (i = 0; i < m; i++) {
      a = Ed[i].first, b = Ed[i].second;
      E[a].push_back(b);
      F[b].push_back(a);
      DegE[b]++;
      DegF[a]++;
    }
    for (i = 1; i <= n; i++)
      if (!DegE[i]) Q[++tail] = i;
    while (head < tail) {
      int x = Q[++head];
      for (auto &y : E[x]) {
        be[y] = max(be[y], be[x] + 1);
        DegE[y]--;
        if (!DegE[y]) Q[++tail] = y;
      }
    }
    if (tail != n) {
      printf("-1\n");
      return;
    }
    head = tail = 0;
    for (i = 1; i <= n; i++)
      if (!DegF[i]) Q[++tail] = i;
    while (head < tail) {
      int x = Q[++head];
      for (auto &y : F[x]) {
        ed[y] = min(ed[y], ed[x] - 1);
        DegF[y]--;
        if (!DegF[y]) Q[++tail] = y;
      }
    }
    for (i = 1; i <= n; i++) {
      if (be[i] > ed[i]) {
        printf("-1\n");
        return;
      }
      G[be[i]].push_back(i);
    }
    priority_queue<pair<int, int> > PQ;
    for (i = 1; i <= n; i++) {
      for (auto &tp : G[i]) {
        PQ.push({-ed[tp], tp});
      }
      if (PQ.empty()) {
        printf("-1\n");
        return;
      }
      int x = PQ.top().second;
      PQ.pop();
      if (ed[x] < i) {
        printf("-1\n");
        return;
      }
      Res[x] = i;
    }
    for (i = 1; i <= n; i++) RRes[Res[i]] = i;
  }
} DD;
int n, m, w[300][300], X[300], Y[300], C[101000], R[300][300], vis[101000];
int A[300], B[300];
vector<int> E[101000];
struct point {
  int x, y, c;
  bool operator<(const point &p) const { return c < p.c; }
};
int Num(int x, int y) { return (x - 1) * m + y; }
void Solve() {
  int i, j;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      scanf("%d", &w[i][j]);
      X[i] = max(X[i], w[i][j]);
      Y[j] = max(Y[j], w[i][j]);
    }
  }
  for (i = 1; i <= n; i++) C[X[i]] |= 1;
  for (i = 1; i <= m; i++) C[Y[i]] |= 2;
  int cnt = 0;
  for (i = 1; i <= n * m; i++) {
    if (C[i] == 3) {
      vis[i] = 1;
      cnt++;
      R[cnt][cnt] = i;
      A[cnt] = B[cnt] = i;
    }
  }
  int cx = cnt, cy = cnt;
  for (i = n * m; i >= 1; i--) {
    if (C[i] == 1) {
      cx++;
      R[cx][cnt] = i;
      vis[i] = 1;
      A[cx] = i;
    }
    if (C[i] == 2) {
      cy++;
      R[cnt][cy] = i;
      vis[i] = 1;
      B[cy] = i;
    }
  }
  for (i = 1; i < cnt; i++) {
    for (j = i; j < m; j++) {
      E[Num(i, j)].push_back(Num(i, j + 1));
    }
    for (j = i; j < n; j++) {
      E[Num(j, i)].push_back(Num(j + 1, i));
    }
    for (j = i + 1; j <= m; j++) {
      E[Num(i + 1, j)].push_back(Num(i, j));
    }
    for (j = i + 1; j <= n; j++) {
      E[Num(j, i + 1)].push_back(Num(j, i));
    }
  }
  for (i = cnt; i <= n; i++) {
    for (j = cnt; j <= m; j++) {
      if (i != n) E[Num(i, j)].push_back(Num(i + 1, j));
      if (j != m) E[Num(i, j)].push_back(Num(i, j + 1));
    }
  }
  int pv = 0;
  vector<pair<int, int> > Ran;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      if (R[i][j])
        Ran.push_back({R[i][j], R[i][j]});
      else
        Ran.push_back({1, n * m});
    }
  }
  vector<pair<int, int> > Ed;
  for (i = 1; i <= n * m; i++) {
    for (auto &t : E[i]) {
      Ed.push_back({t, i});
    }
  }
  DD.Do(Ed, Ran);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      printf("%d ", DD.Res[Num(i, j)]);
    }
    puts("");
  }
}
int main() {
  int TC = 1;
  while (TC--) {
    Solve();
  }
}