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>

#define SZ(x) (((int)x.size()))

using namespace std;

const int maxn = 110;
int n, m, val, a[maxn][maxn], mat[maxn];
bool mark[maxn];
vector<int> vec[maxn][maxn], tmp;
unordered_set<int> adj[maxn];

bool dfs(int x)
{
	if (mark[x])
		return 0;
	mark[x] = 1;
	for (auto u : adj[x])
	{
		if (mat[u] == -1 || dfs(mat[u]))
		{
			mat[u] = x;
			return 1;
		}
	}
	return 0;
}

int main()
{
	cin >> n >> m;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cin >> val;
			val--;
			vec[i][val / m].push_back(val + 1);
			adj[val / m].insert(i);
		}
	}
	for (int i = 0; i < m; i++)
	{
		memset(mat, -1, sizeof mat);
		for (int j = 0; j < n; j++)
		{
			memset(mark, 0, sizeof mark);
			dfs(j);
		}
		for (int j = 0; j < n; j++)
		{
			a[j][i] = vec[j][mat[j]].back();
			vec[j][mat[j]].pop_back();
			if (SZ(vec[j][mat[j]]) == 0)
				adj[mat[j]].erase(j);
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
			cout << a[i][j] << " ";
		cout << endl;
	}
	for (int i = 0; i < m; i++)
	{
		tmp.clear();
		for (int j = 0; j < n; j++)
			tmp.push_back(a[j][i]);
		sort(tmp.begin(), tmp.end());
		for (int j = 0; j < n; j++)
			a[j][i] = tmp[j];
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
			cout << a[i][j] << " ";
		cout << endl;
	}
	return 0;
}