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;
class matching{
  public:
  vector< vector<int> > g;
  vector<int> pa;
  vector<int> pb;
  vector<int> was;
  int n, m;
  int res;
  int iter;
  matching(int _n, int _m) : n(_n), m(_m) {
    assert(0 <= n && 0 <= m);
    pa = vector<int>(n, -1);
    pb = vector<int>(m, -1);
    was = vector<int>(n, 0);
    g.resize(n);
    res = 0;
    iter = 0;
  }
  void add(int from, int to) {
    assert(0 <= from && from < n && 0 <= to && to < m);
    g[from].push_back(to);
  }
  bool dfs(int v) {
    was[v] = iter;
    for (int u : g[v]) {
      if (pb[u] == -1) {
        pa[v] = u;
        pb[u] = v;
        return true;
      }
    }
    for (int u : g[v]) {
      if (was[pb[u]] != iter && dfs(pb[u])) {
        pa[v] = u;
        pb[u] = v;
        return true;
      }
    }
    return false;
  }
  int solve() {
    while (true) {
      iter++;
      int add = 0;
      for (int i = 0; i < n; i++) {
        if (pa[i] == -1 && dfs(i)) {
          add++;
        }
      }
      if (add == 0) {
        break;
      }
      res += add;
    }
    return res;
  }
  int run_one(int v) {
    if (pa[v] != -1) {
      return 0;
    }
    iter++;
    return (int) dfs(v);
  }
};
int n,m;
int a[101][101];
int main(){
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=m ;j++){
			cin >> a[i][j];
		}
	}
	for(int i=1; i<=m ;i++){
		matching g(n+1,n+1);
		for(int j=i; j<=m ;j++){
			for(int k=1; k<=n ;k++){
				g.add(k,(a[k][j]-1)/m+1);
			}
		}
		if(g.solve()!=n) return 49;
		for(int k=1; k<=n ;k++){
			for(int j=i; j<=m ;j++){
				if(g.pa[k]==(a[k][j]-1)/m+1){
					swap(a[k][i],a[k][j]);
					break;
				}
			}
		}
	}
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=m ;j++){
			cout << a[i][j] << ' ';
		}
		cout << '\n';
	}
	int t[101];
	for(int i=1; i<=m ;i++){
		for(int j=1; j<=n ;j++){
			t[j]=a[j][i];
		}
		sort(t+1,t+n+1);
		for(int j=1; j<=n ;j++){
			a[j][i]=t[j];
		}
	}
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=m ;j++){
			cout << a[i][j] << ' ';
		}
		cout << '\n';
	}
}