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 long ll;

#ifdef iq
  mt19937 rnd(228);
#else
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif

vector <int> gr[100];
bool u[100];
int mt[100];
vector <int> gays[100][100];

bool dfs(int v) {
  if (u[v]) {
    return false;
  }
  u[v] = true;
  for (int to : gr[v]) {
    if (gays[v][to].empty()) continue;
    if (mt[to] == -1 || dfs(mt[to])) {
      mt[to] = v;
      return true;
    }
  }
  return false;
}

int main() {
#ifdef iq
  freopen("a.in", "r", stdin);
#endif
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector <int> ind(n * m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      ind[i * m + j] = i;
    }
  }
  vector <vector <int> > a(n, vector <int> (m));
  for (int i = 0; i < n; i++) mt[i] = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      x--;
      a[i][j] = x;
      gr[ind[x]].push_back(i);
      gays[ind[x]][i].push_back(x);
    }
  }
  vector <vector <int> > ans(n);
  for (int it = 0; it < m; it++) {
    for (int i = 0; i < n; i++) mt[i] = -1, u[i] = 0;
    for (int i = 0; i < n; i++) {
      assert(dfs(i));
      for (int j = 0; j < n; j++) u[j] = 0;
    }
    for (int i = 0; i < n; i++) {
      ans[i].push_back(gays[mt[i]][i].back());
      gays[mt[i]][i].pop_back();
    }
  }
  vector <vector <pair <int, int> > > st(m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      st[j].push_back({ind[ans[i][j]], ans[i][j]});
      cout << ans[i][j] + 1 << ' ';
    }
    cout << '\n';
  }
  for (int i = 0; i < m; i++) {
    sort(st[i].begin(), st[i].end());
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cout << st[j][i].second + 1 << ' ';
    }
    cout << '\n';
  }
}