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>
struct MI {
 private:
  char bb[1 << 14];
  FILE *f;
  char *bs, *be;
  char e;
  bool o, l;

 public:
  MI() : f(stdin), bs(0), be(0) {}
  inline char get() {
    if (o) {
      o = 0;
      return e;
    }
    if (bs == be) be = (bs = bb) + fread(bb, 1, sizeof(bb), f);
    if (bs == be) {
      l = 1;
      return -1;
    };
    return *bs++;
  }
  inline void unget(char c) {
    o = 1;
    e = c;
  }
  template <class T>
  inline MI &operator>(T &);
};
template <class T>
struct Q {
  const static bool U = T(-1) >= T(0);
  inline void operator()(MI &t, T &r) const {
    r = 0;
    char c;
    bool y = 0;
    if (U)
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (isdigit(c)) break;
      }
    else
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (c == '-') {
          c = t.get();
          if (isdigit(c)) {
            y = 1;
            break;
          };
        } else if (isdigit(c))
          break;
        ;
      };
    for (;;) {
      if (c == -1) goto E;
      if (isdigit(c))
        r = r * 10 + (c ^ 48);
      else
        break;
      c = t.get();
    }
    t.unget(c);
  E:;
    if (y) r = -r;
  }
};
template <>
struct Q<char> {};
template <class T>
inline MI &MI::operator>(T &t) {
  Q<T>()(*this, t);
  return *this;
}
template <class T>
std::ostream &operator<(std::ostream &out, const T &t) {
  return out << t;
}
using std::cout;
MI cin;
const int $n = 255;
template <typename T>
inline bool gmax(T &a, const T &b) {
  return a < b && (a = b, true);
}
int n, m, v[$n][$n], mh[$n], mv[$n];
bool eh[$n * $n], ev[$n * $n];
int main() {
  cin > n > m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      cin > v[i][j];
      gmax(mh[i], v[i][j]);
      gmax(mv[j], v[i][j]);
    }
  for (int i = 1; i <= n; ++i) eh[mh[i]] = 1;
  for (int i = 1; i <= m; ++i) ev[mv[i]] = 1;
  std::queue<std::pair<int, int>> q;
  int tx = 0, ty = 0;
  for (int i = n * m; i; --i) {
    if (eh[i]) ++tx;
    if (ev[i]) ++ty;
    if (eh[i] || ev[i])
      v[tx][ty] = i;
    else {
      auto [x, y] = q.front();
      q.pop();
      v[x][y] = i;
    }
    if (eh[i])
      for (int j = ty - 1; j; --j) q.emplace(tx, j);
    if (ev[i])
      for (int j = tx - 1; j; --j) q.emplace(j, ty);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cout < v[i][j] < " \n"[j == m];
}