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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <vector>
using namespace std;

namespace zyt
{
	template<typename T>
	inline bool read(T &x)
	{
		char c;
		bool f = false;
		x = 0;
		do
			c = getchar();
		while (c != EOF && c != '-' && !isdigit(c));
		if (c == EOF)
			return false;
		if (c == '-')
			f = true, c = getchar();
		do
			x = x * 10 + c - '0', c = getchar();
		while (isdigit(c));
		if (f)
			x = -x;
		return true;
	}
	template<typename T>
	inline void write(T x)
	{
		static char buf[20];
		char *pos = buf;
		if (x < 0)
			putchar('-'), x = -x;
		do
			*pos++ = x % 10 + '0';
		while (x /= 10);
		while (pos > buf)
			putchar(*--pos);
	}
	const int N = 110;
	int n, m, arr[N][N], b[N][N];
	vector<int> g[N];
	namespace Hungary
	{
		int match[N], vis[N], time;
		bool find(const int u)
		{
			for (auto v : g[u])
				if (vis[v] < time)
				{
					vis[v] = time;
					if (!match[v] || find(match[v]))
					{
						match[v] = u;
						return true;
					}
				}
			return false;
		}
		void solve()
		{
			time = 0;
			memset(match, 0, sizeof(int[n + 1]));
			memset(vis, 0, sizeof(int[n + 1]));
			for (int i = 1; i <= n; i++)
				++time, find(i);
		}
	}
	bool used[N][N];
	int work()
	{
		using namespace Hungary;
		read(n), read(m);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				read(arr[i][j]);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				g[i].push_back((arr[i][j] - 1) / m + 1);
		for (int i = 1; i <= m; i++)
		{
			solve();
			for (int j = 1; j <= n; j++)
			{
				for (int k = 1; k <= m; k++)
					if (!used[match[j]][k] && (arr[match[j]][k] - 1) / m + 1 == j)
					{
						used[match[j]][k] = true, b[match[j]][i] = arr[match[j]][k];
						break;
					}
				for (auto it = g[j].begin(); it != g[j].end(); it++)
					if (match[*it] == j)
					{
						g[j].erase(it);
						break;
					}
			}
		}
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
				printf("%d ", b[i][j]);
			putchar('\n');
		}
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
				for (int k = 1; k <= n; k++)
					if ((b[k][j] - 1) / m + 1 == i)
					{
						printf("%d ", b[k][j]);
						break;
					}
			putchar('\n');
		}
		return 0;
	}
}
int main()
{
	return zyt::work();
}