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;
typedef long double ld;
typedef long long ll;
typedef pair<double, double> pdd;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef vector<pii> vii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
//const int mod = ;

void out(vvi v) {
  for (int i = 0; i < v.size(); ++i) {
    for (int x : v[i]) cout << x + 1 << ' ';
    cout << endl;
  }
}

vvi gr;
vector<int> mt;
vector<char> used; 
bool try_kuhn (int v) {
  if (used[v])  return false;
  used[v] = true;
  for (size_t i=0; i<gr[v].size(); ++i) {
    int to = gr[v][i];
    if (mt[to] == -1 || try_kuhn (mt[to])) {
      mt[to] = v;
      mt[v] = to;
      return true;
    }
  }
  return false;
}

int main() {
  std::ios::sync_with_stdio(false); std::cin.tie(0);
  int n, m;
  cin >> n >> m;
  vvi a(n, vi(m));
  vector<vvi> ra(n, vvi(n));
  for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) {
    cin >> a[i][j];
    --a[i][j];
    ra[i][a[i][j] / m].push_back(a[i][j]);
  }
  vvi b(n, vi(m));
  vvi c = b;
  for (int col = 0; col < m; ++col) {
    gr.assign(2 * n, vi());
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) if (!ra[i][j].empty()) {
        gr[i].push_back(n + j);
      }
    }
    mt.assign (gr.size(), -1);
    for (int run = 1; run;) {
      run = 0;
      used.assign(gr.size(), false);
      for (int v=0; v<n; ++v) if (mt[v] == -1 && try_kuhn(v)) {
        run = 1;
      }
    }
    for (int i = 0; i < n; ++i) {
      assert(mt[i] != -1);
      int j = mt[i] - n;
      b[i][col] = ra[i][j].back();
      ra[i][j].pop_back();
    }
  }
  for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) {
    c[b[i][j] / m][j] = b[i][j];
  }
  out(b);
  out(c);
  return 0;
}