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;


using ll = long long;


mt19937 mt(736);


bool dfs(const vector<vector<int>> &gr, vector<bool> &used, vector<int> &rd, int v)
{
	if (used[v])
		return false;
	used[v] = true;

	for (auto it : gr[v])
		if (rd[it] == -1 || dfs(gr, used, rd, rd[it]))
		{
			rd[it] = v;

			return true;
		}
}


void Kuhn(const vector<vector<int>> &gr, vector<int> &rd)
{
	vector<bool> matched(gr.size());

	bool loop = true;

	while (loop)
	{
		loop = false;

		vector<bool> used(gr.size());

		for (int i = 0; i < gr.size(); i++)
			if (!matched[i])
				loop |= matched[i] = dfs(gr, used, rd, i);
	}
}


void solve(istream &cin = std::cin, ostream &cout = std::cout)
{
	int n, m;

	cin >> n >> m;

	vector<vector<int>> matr(n, vector<int>(m));

	for (auto &row : matr)
		for (auto &it : row)
		{
			cin >> it;
			it--;
		}


	vector<vector<int>> gr(n);


	for (int r = 0; r < n; r++)
		for (auto it : matr[r])
			gr[r].push_back(it / m);


	vector<vector<int>> transpone(m, vector<int> (n));

	for (auto &col : transpone)
	{
		vector<int> rd(n, -1);

		Kuhn(gr, rd);

		for (int i = 0; i < rd.size(); i++)
		{
			col[rd[i]] = i;

			gr[rd[i]].erase(find(gr[rd[i]].begin(), gr[rd[i]].end(), i));
		}
	}

	for (int r = 0; r < n; r++)
		for (int c = 0; c < m; c++)
		{
			int pos = 0;

			while (pos < matr[r].size() && matr[r][pos] / m != transpone[c][r])
				pos++;

			assert(pos < matr[r].size());

			transpone[c][r] = matr[r][pos];

			swap(matr[r][pos], matr[r].back());
			matr[r].pop_back();
		}


	for (int i = 0; i < n; i++)
	{
		for (const auto &row : transpone)
			cout << row[i] + 1 << ' ';
		cout << endl;
	}

	for (auto &row : transpone)
		sort(row.begin(), row.end());

	for (int i = 0; i < n; i++)
	{
		for (const auto &row : transpone)
			cout << row[i] + 1 << ' ';
		cout << endl;
	}
}


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	cout << fixed;

#ifdef LOCAL
	auto st = clock();

	ifstream fin("../input.txt");

	solve(fin);

	cout << "clock: " << setprecision(4) << (clock() - st) / (double) CLOCKS_PER_SEC << endl;
#else
	solve();
#endif

	return 0;
}