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;
#ifdef DEBUG
	#define D(x...) printf(x)
#else
	#define D(x...)
#endif
int amperrow[110][110], n, m, upto, seen[210], ismatched[210], matchedto[210];
vector<int> adj[210];
int match(int a)
{
	if (seen[a] == upto) return 0;
	seen[a] = upto;
	for (auto b : adj[a])
	{
		if (!ismatched[b])
		{
			ismatched[b] = ismatched[a] = 1;
			matchedto[b] = a;
			return 1;
		}
		else if (match(matchedto[b]))
		{
			ismatched[a] = 1;
			matchedto[b] = a;
			return 1;
		}
	}
	return 0;
}
int moved[110][110], val[110][110];
pair<int, int> movedto[110][110];
int rowitwants(int a)
{
	a--;
	return a/m;
}	
void dobpmthing(int jmovingto)
{
	D("starting new %d\n", jmovingto);
	for (int i = 0; i < n+n; i++) adj[i].clear();
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (amperrow[i][j]) adj[i].push_back(n+j), adj[n+j].push_back(i);
			D("%d ", amperrow[i][j]);
		}
		D("\n");
	}
	fill_n(ismatched, 2*n, 0);
	int ans = 0;
	for (int i = 0; i < n; i++)
	{
		upto++;
		if (match(i))
		{
			ans++;
			D("yay %d\n", i);
		}
		else
		{
			D("rip %d\n", i);
		}
	}
	assert(ans == n);
	for (int i = n; i < 2*n; i++)
	{
		bool done = 0;
		for (int j = 0; j < m; j++)
		{
			if (rowitwants(val[matchedto[i]][j]) == i-n && !moved[matchedto[i]][j])
			{
				done = 1;
				moved[matchedto[i]][j] = 1;
				movedto[matchedto[i]][j] = { matchedto[i], jmovingto };
				break;
			}
		}
		assert(done);
		amperrow[matchedto[i]][i-n]--;
	}
}
int nowhere[110][110];
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			scanf("%d", &val[i][j]);
			amperrow[i][rowitwants(val[i][j])]++;
		}
	}
	for (int i = 0; i < m; i++)
	{
		dobpmthing(i);
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			nowhere[movedto[i][j].first][movedto[i][j].second] = val[i][j];
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++) printf("%d ", nowhere[i][j]);
		printf("\n");
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			int done = 0;
			for (int k = 0; k < n; k++)
			{
				if (rowitwants(nowhere[k][j]) == i) printf("%d ", nowhere[k][j]), done++;
			}
			assert(done == 1);
		}
		printf("\n");
	}
}