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;
const long long MAXN = 300;
long long A[MAXN][MAXN];
long long res[MAXN][MAXN];
bool mk[MAXN][MAXN];
long long ingree[MAXN][MAXN];
vector<pair<long long, long long>> L[MAXN][MAXN];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long N, M;
  cin >> N >> M;
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < M; j++) {
      cin >> A[i][j];
    }
  }
  set<long long> st1;
  set<long long> st2;
  for (long long i = 0; i < N; i++) {
    long long may = 0;
    for (long long j = 0; j < M; j++) {
      may = max(may, A[i][j]);
    }
    st1.insert(may);
  }
  for (long long j = 0; j < M; j++) {
    long long may = 0;
    for (long long i = 0; i < N; i++) {
      may = max(may, A[i][j]);
    }
    st2.insert(may);
  }
  long long px = -1, py = -1;
  set<pair<long long, pair<long long, long long>>> st;
  for (long long i = N * M; i >= 1; i--) {
    bool f1 = false, f2 = false;
    if (st1.find(i) != st1.end()) px++, f1 = true;
    if (st2.find(i) != st2.end()) py++, f2 = true;
    if (f2) {
      for (long long x = 0; x < N; x++) {
        if (x == px) continue;
        if (x < px) {
          ingree[x][py]++;
          L[x + 1][py].push_back({x, py});
        }
        if (x > px) {
          ingree[x][py]++;
          L[x - 1][py].push_back({x, py});
        }
      }
      for (long long x = 0; x < px; x++) {
        st.insert({ingree[x][py], {x, py}});
        mk[x][py] = true;
      }
    }
    if (f1) {
      for (long long y = 0; y < M; y++) {
        if (y == py) continue;
        if (y < py) {
          ingree[px][y]++;
          L[px][y + 1].push_back({px, y});
        }
        if (y > py) {
          ingree[px][y]++;
          L[px][y - 1].push_back({px, y});
        }
      }
      for (long long y = 0; y < py; y++) {
        st.insert({ingree[px][y], {px, y}});
        mk[px][y] = true;
      }
    }
    if (f1 || f2) {
      res[px][py] = i;
      mk[px][py] = true;
      for (auto v : L[px][py]) {
        if (mk[v.first][v.second]) {
          st.erase({ingree[v.first][v.second], {v.first, v.second}});
          ingree[v.first][v.second]--;
          st.insert({ingree[v.first][v.second], {v.first, v.second}});
        } else {
          ingree[v.first][v.second]--;
        }
      }
    } else {
      pair<long long, long long> pp = (*st.begin()).second;
      st.erase(st.begin());
      res[pp.first][pp.second] = i;
      for (auto v : L[pp.first][pp.second]) {
        if (mk[v.first][v.second]) {
          st.erase({ingree[v.first][v.second], {v.first, v.second}});
          ingree[v.first][v.second]--;
          st.insert({ingree[v.first][v.second], {v.first, v.second}});
        } else {
          ingree[v.first][v.second]--;
        }
      }
    }
  }
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < M; j++) {
      cout << res[i][j] << " ";
    }
    cout << '\n';
  }
  return 0;
}