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 <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 105
int n, m, linky[N], mat[N][N];
bool A[N][N], vis[N];
vector<int>vec[N][N];
bool Dfs(int x)
{
	for (int i = 1; i <= n; i++)
	{
		if (!vis[i] && A[x][i])
		{
			vis[i] = 1;
			if (!linky[i] || Dfs(linky[i]))
			{
				linky[i] = x;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			int x;
			scanf("%d",&x);
			vec[i][(x - 1) / m + 1].push_back(x);
		}
	}
	for (int i = 1; i <= m; i++)
	{
		for (int j = 1; j <= n; j++)
			for (int k = 1; k <= n; k++)
				A[j][k] = (!vec[j][k].empty());
		memset(linky, 0, sizeof(linky));
		for (int j = 1; j <= n; j++)
		{
			memset(vis, 0, sizeof(vis));
			Dfs(j);
		}
		for (int j = 1; j <= n; j++)
		{
			int k = linky[j];
			mat[k][i] = vec[k][j].back();
			vec[k][j].pop_back();
		}
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			printf("%d%c",mat[i][j], j == m ? 10 : 32);
	for (int i = 1; i <= m; i++)
		for (int j = 1; j <= n; j++)
			for (int k = j + 1; k <= n; k++)
				if (mat[j][i] > mat[k][i])
					swap(mat[j][i], mat[k][i]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			printf("%d%c", mat[i][j], j == m ? 10 : 32);
}