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;
const int N = 110;
int n,m,bel[N],a[N][N];
bool E[N][N],vis[N];
vector<int> vec[N][N];
bool find(int i)
{
	for(int j = 1;j <= n;j ++)
		if(!vis[j] && E[i][j])
		{
			vis[j] = 1;
			if(!bel[j] || find(bel[j]))
			{
				bel[j] = i;
				return 1;
			}
		}
	return 0;
}
void print()
{
	for(int i = 1;i <= n;i ++)
		for(int j = 1;j <= m;j ++)
			printf("%d%c",a[i][j],j == m?'\n':' ');
}
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 ++)
				E[j][k] = (!vec[j][k].empty());
		memset(bel,0,sizeof(bel));
		for(int j = 1;j <= n;j ++)
			memset(vis,0,sizeof(vis)),find(j);
		for(int j = 1;j <= n;j ++)
		{
			int k = bel[j];
			a[k][i] = vec[k][j].back();
			vec[k][j].pop_back();
		}
	}
	print();
	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] > a[k][i])
					swap(a[j][i],a[k][i]);
	print();
	return 0;
}