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 <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
 
using namespace std;
 
typedef long long ll;
 
const ll INF = 1e9, MOD = 1e9 + 7, MOD2 = 1e6 + 3;

int p[1000000], u[100], n, m;

vector <int> g[1000000];

int v[100][100], a[100][100];

bool kuhn (int x)
{
	if (u[x]) return false;
	u[x] = 1;

	for (int to : g[x])
		if (p[to] == -1 || kuhn (p[to]))
		{
			p[to] = x;
			return true;
		}

	return false;
}

int main ()
{
	cin >> n >> m;

	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			scanf ("%d", &a[i][j]);

	for (int k = 0; k < m; k++)
	{
		memset (p, -1, sizeof (p));

		for (int i = 0; i < n; i++)
		{
			g[i].clear ();
			for (int j = k; j < m; j++)
			{
				int d = (a[i][j] - 1) / m;
				g[i].push_back (d);
			}
		}

		for (int i = 0; i < n; i++)
		{
			kuhn (i);
			memset (u, 0, sizeof (u));
		}

		vector <int> to (n);

		for (int i = 0; i < n; i++)
			to[p[i]] = i;

		for (int i = 0; i < n; i++)
			for (int j = k; j < m; j++)
			{
				if ((a[i][j] - 1) / m == to[i])
				{
					swap (a[i][k], a[i][j]);
					break;
				}
			}
	}

	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
			printf ("%d ", a[i][j]);
		putchar ('\n');
	}

	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			v[(a[i][j] - 1) / m][j] = a[i][j];

	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
			printf ("%d ", v[i][j]);
		putchar ('\n');
	}
}