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;
template <typename T>
bool vmin(T& a, T b) {
  return (a > b) ? (a = b, true) : (false);
}
template <typename T>
bool vmax(T& a, T b) {
  return (a < b) ? (a = b, true) : (false);
}
template <typename T>
T smax(T x) {
  return x;
}
template <typename T, typename... K>
T smax(T a, const K&... args) {
  return max(a, smax(args...));
}
template <typename T>
T smin(T x) {
  return x;
}
template <typename T, typename... K>
T smin(T a, const K&... args) {
  return min(a, smin(args...));
}
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& z) {
  os << "(" << z.first << ", " << z.second << ')';
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& a) {
  bool isfirst = true;
  os << "{";
  for (auto z : a) {
    if (!isfirst) {
      os << ", ";
    }
    os << z;
    isfirst = false;
  }
  os << '}';
  return os;
}
const int inf = (signed)(~0u >> 2);
const long long llf = (signed long long)(~0ull >> 2);
template <typename T>
int vsize(vector<T>& x) {
  return (signed)x.size();
}
template <typename T>
int discrete(T* a, int* b, int n) {
  vector<T> v(a + 1, a + n + 1);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; i++)
    b[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  return v.size();
}
mt19937 rng(time(NULL));
int randint(int l, int r) { return rng() % (r - l + 1) + l; }
typedef class Data {
 public:
  int x, y;
  Data() {}
  Data(int x, int y) : x(x), y(y) {}
} Data;
const int N = 255;
int n, m;
int w, h;
queue<Data> Q;
int mxc[N], mxr[N];
bool inq[N][N];
int a[N][N], b[N][N];
bool iscmx[N * N], isrmx[N * N];
bool chk(int x, int y) {
  if (x >= 1 && y >= 1 && x <= w && y <= h && !b[x][y] && !inq[x][y]) {
    if (!b[x][y + (mxr[x] < y ? -1 : 1)]) return false;
    if (!b[x + (mxc[y] < x ? -1 : 1)][y]) return false;
    inq[x][y] = true;
    Q.push(Data(x, y));
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", a[i] + j);
    }
  }
  for (int i = 1; i <= n; i++) {
    int mx = 0;
    for (int j = 1; j <= m; j++) {
      vmax(mx, a[i][j]);
    }
    isrmx[mx] = true;
  }
  for (int j = 1; j <= m; j++) {
    int mx = 0;
    for (int i = 1; i <= n; i++) {
      vmax(mx, a[i][j]);
    }
    iscmx[mx] = true;
  }
  for (int i = n * m; i; i--) {
    if (isrmx[i] && iscmx[i]) {
      ++w, ++h;
      b[w][h] = i;
      mxr[w] = h;
      mxc[h] = w;
      chk(w - 1, h);
      chk(w, h - 1);
    } else if (isrmx[i]) {
      ++w;
      b[w][h] = i;
      mxr[w] = h;
      chk(w, h - 1);
    } else if (iscmx[i]) {
      ++h;
      b[w][h] = i;
      mxc[h] = w;
      chk(w - 1, h);
    } else {
      assert(!Q.empty());
      int x = Q.front().x, y = Q.front().y;
      Q.pop();
      b[x][y] = i;
      chk(x - 1, y);
      chk(x, y - 1);
      chk(x + 1, y);
      chk(x, y + 1);
    }
  }
  assert(w == n && h == m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      printf("%d%c", b[i][j], (j == m) ? '\n' : ' ');
    }
  }
  return 0;
}