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:
/*input
3 2
2 6
4 3
1 5
*/
#include <bits/stdc++.h>
#define pb push_back
#define all(a) a.begin(), a.end()
#define sz(a) (int)a.size()
#define x first
#define y second
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int maxn = 205;
const int mod = 1e9 + 7;
int n, m, arr[maxn][maxn], match[maxn], vis[maxn][maxn], ans[maxn][maxn], mark[maxn];
multiset<int>adj[maxn];
bool dfs(int v)
{
	if(mark[v])
		return false;
	mark[v] = true;
	for(int to : adj[v])
	{
		if(match[to] == -1 || dfs(match[to]))
		{
			match[v] = to;
			match[to] = v;
			return true;
		}
	}
	return false;
}
int main()
{
	ios_base::sync_with_stdio(false), cin.tie(0);
	cin >> n >> m;
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			cin >> arr[i][j];
			--arr[i][j];
		}
	}
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			adj[i + n].insert(arr[i][j] / m);
		}
	}
	for(int j = 0; j < m; j++)
	{
		memset(match, -1, sizeof match);
		while(true)
		{
			memset(mark, 0, sizeof mark);
			bool chk = false;
			for(int i = n; i < 2 * n; i++)
				if(match[i] == -1 && !mark[i])
					chk |= dfs(i);
			if(!chk)
				break;
		}
		for(int i = 0; i < n; i++)
		{
			int color = match[i + n];
			adj[i + n].erase(adj[i + n].find(color));
			for(int jj = 0; jj < m; jj++)
			{
				if(arr[i][jj] / m == color && !vis[i][jj])
				{
					vis[i][jj] = 1;
					ans[i][j] = arr[i][jj];
					break;
				}
			}
		}
	}
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
			cout << ans[i][j] + 1 << " ";
		cout << "\n";
	}
	for(int j = 0; j < m; j++)
	{
		for(int i = 0; i < n; i++)
		{
			for(int ii = i + 1; ii < n; ii++)
			{
				if(ans[i][j] > ans[ii][j])
					swap(ans[i][j], ans[ii][j]);
			}
		}
	}
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
			cout << ans[i][j] + 1 << " ";
		cout << "\n";
	}
	return 0;
}