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<algorithm>
#include<cstdio>
#include<cstring>
#define M 10001
#define N 101
using namespace std;
int A[N][N],B[M],C[N][N],D[N][N],b[N],d[N],m,n;bool a[N][N],c[N];
bool dfs(int u)
{
	c[u]=true;
	for(int i=1;i<=n;i++)if(a[u][i]&&(!d[i]||!c[d[i]]&&dfs(d[i])))
	{
		b[d[i]=u]=i;
		return true;
	}
	return false;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)for(int j=1,u;j<=m;j++)scanf("%d",&u),B[u]=A[i][(u+m-1)/m],A[i][(u+m-1)/m]=u;
	for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)a[i][j]=A[i][j];
	for(int i=1;i<=m;i++)
	{
		for(int j=1;j<=n;j++)if(!b[j])dfs(j),memset(c+1,0,n); 
		for(int j=1;j<=n;j++)C[j][i]=A[j][b[j]];
		for(int j=1;j<=n;j++)if(!(A[j][b[j]]=B[A[j][b[j]]]))b[j]=d[b[j]]=a[j][b[j]]=0;
	}
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)printf("%d%c",C[i][j],j==m?'\n':' ');
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)D[j][i]=C[i][j];
	for(int i=1;i<=m;i++)sort(D[i]+1,D[i]+n+1);
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)printf("%d%c",D[j][i],j==m?'\n':' ');
	return 0;
}