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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

const int MAXN = 105;

int n,m;
int match[MAXN];
int cnt[MAXN][MAXN];

bool vis[MAXN];

pair<int,int> b[MAXN];
pair<int,int> a[MAXN][MAXN];

bool dfs(int i)
{
	if (vis[i])
		return 0;
	vis[i] = 1;
	for (int j = 1;j <= n;j++)
		if (cnt[j][i] && (!match[j] || dfs(match[j])))
		{
			match[j] = i;
			vis[i] = 0;
			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++)
		{
			scanf("%d",&a[i][j].first);
			a[i][j].second = (a[i][j].first - 1) % m + 1;
			a[i][j].first = (a[i][j].first - 1) / m + 1;
			cnt[i][a[i][j].first]++;
		}
	for (int i = 1;i <= m;i++)
	{
		memset(vis,0,sizeof(vis));
		memset(match,0,sizeof(match));
		for (int j = 1;j <= n;j++)
			dfs(j);
		for (int j = 1;j <= n;j++)
		{
			cnt[j][match[j]]--;
			for (int k = i;k <= m;k++)
				if (a[j][k].first == match[j])
				{
					swap(a[j][k],a[j][i]);
					break;
				}
		}
	}
	for (int i = 1;i <= n;i++,printf("\n"))
		for (int j = 1;j <= m;j++)
			printf("%d ",(a[i][j].first - 1) * m + a[i][j].second);
	for (int i = 1;i <= m;i++)
		for (int j = 1;j <= n;j++)
			for (int k = j + 1;k <= n;k++)
				if (a[j][i].first > a[k][i].first)
					swap(a[j][i],a[k][i]);
	for (int i = 1;i <= n;i++,printf("\n"))
		for (int j = 1;j <= m;j++)
			printf("%d ",(a[i][j].first - 1) * m + a[i][j].second);
	return 0;
}