Problem: We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.

You need to rearrange these numbers as follows:

1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.



After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.

Constraints

* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.

Input

Input is given from Standard Input in the following format:


N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}


Output

Print one way to rearrange the numbers in the following format:


B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}


Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.

Examples

Input

3 2
2 6
4 3
1 5


Output

2 6
4 3
5 1
2 1
4 3
5 6


Input

3 4
1 4 7 10
2 5 8 11
3 6 9 12


Output

1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <string>
#include <random>
#include <cassert>
#include <cstring>
#include <chrono>

using namespace std;

typedef long long ll;

int n, m;
int a[100][100];
int b[100][100];
int ansA[100][100];
int ansB[100][100];
int rowInA[100 * 100];
int rowInB[100 * 100];

vector<int> g[100][100];
bool used[100];
int R[100];

bool dfs(int v) {
  if (used[v]) return false;
  used[v] = true;
  for (int to = 0; to < n; ++to) {
    if (g[v][to].empty()) continue;
    if (R[to] == -1 || dfs(R[to])) {
      R[to] = v;
      return true;
    }
  }
  return false;
}

int main()
{
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> a[i][j];
      a[i][j]--;
      rowInA[a[i][j]] = i;
    }
  }

  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      b[i][j] = i * m + j;
      rowInB[b[i][j]] = i;
    }
  }

  for (int i = 0; i < n * m; ++i) {
    g[rowInA[i]][rowInB[i]].push_back(i);
  }

  for (int step = 0; step < m; ++step) {
    memset(R, -1, sizeof(R));
    int cur = 0;
    for (int row = 0; row < n; ++row) {
      memset(used, false, sizeof(used));
      if (dfs(row)) {
        cur++;
      }
    }
    assert(cur == n);
    for (int i = 0; i < n; ++i) {
      int v = R[i];
      int w = i;
      ansA[v][step] = g[v][w].back();
      ansB[w][step] = g[v][w].back();
      g[v][w].pop_back();
    }
  }

  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (j != 0) cout << " ";
      cout << ansA[i][j] + 1;
    }
    cout << endl;
  }

  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (j != 0) cout << " ";
      cout << ansB[i][j] + 1;
    }
    cout << endl;
  }

  return 0;
}