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 <bits/stdc++.h>

using namespace std;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

struct maximum_bipartite_matching {
  int nl, nr, mbm = 0;
  vector<bool> v;
  vector<int> ml, mr;
  vector<vector<int>>& e;
  maximum_bipartite_matching(int nl, int nr, vector<vector<int>> & e) : nl(nl), nr(nr), ml(nl, -1), mr(nr, -1), e(e) {
    int prv = 0;
    do {
      prv = mbm;
      v.assign(nr, false);
      for (int i = 0; i < nl; i++)
        if (ml[i] == -1)
          mbm += findPath(i);
    } while (mbm > prv);
  }
  bool findPath(int i) {
    for (int j : e[i])
      if (!v[j]) {
        v[j] = true;
        if (mr[j] == -1 || findPath(mr[j])) {
          ml[i] = j; mr[j] = i;
          return true;
        }
      }
    return false;
  }
};

int n, m, a[200][200], b[200][200], c[200][200];
bool u[200][200];

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      cin >> a[i][j], a[i][j]--;
  for (int k = 0; k < m; k++) {
    vector<vector<int>> e(n);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (!u[i][j])
          e[i].push_back(a[i][j] / m);
    maximum_bipartite_matching mbm(n, n, e);
    assert(mbm.mbm == n);
    for (int i = 0; i < n; i++) {
      int j = 0;
      while (u[i][j] || a[i][j] / m != mbm.ml[i])
        j++;
      u[i][j] = true;
      b[i][k] = a[i][j];
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      c[b[i][j] / m][j] = b[i][j];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      cout << b[i][j] + 1 << (j < m - 1 ? " " : "\n");
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      cout << c[i][j] + 1 << (j < m - 1 ? " " : "\n");
}