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;

bool e[105][105],vis[105];
int link[105];

bool dfs(int x,int n) {
  for(int i=1;i<=n;i++)
    if (e[x][i]&&!vis[i]) {
    	vis[i]=1;
    	if (!link[i]||dfs(link[i],n)) {
    		link[i]=x;
    		return 1;
		}
	}
  return 0;
}

void match(int n) {
  memset(link,0,sizeof(link));
  for(int i=1;i<=n;i++) {
  	memset(vis,0,sizeof(vis));
  	dfs(i,n);
  }
}

int a[105][105],b[105][105],val[105][105];
bool use[105][105];

int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++) {
	  scanf("%d",&a[i][j]);
	  val[i][j]=(a[i][j]-1)/m+1;
    }
  for(int i=1;i<=m;i++) {
  	memset(e,0,sizeof(e));
  	for(int j=1;j<=n;j++)
  	  for(int k=1;k<=m;k++)
  	    if (!use[j][k]) e[j][val[j][k]]=1;
  	match(n);
  	for(int j=1;j<=n;j++) {
  		int x=link[j];
  		for(int k=1;k<=m;k++)
  		  if (!use[x][k]&&val[x][k]==j) {
  		  	  use[x][k]=1;
  		  	  b[x][i]=a[x][k];
  		  	  break;
			}
	  }
  }
  for(int i=1;i<=n;i++) {
  	for(int j=1;j<=m;j++) printf("%d ",b[i][j]);
  	printf("\n");
  }
  for(int i=1;i<=m;i++) {
  	for(int j=1;j<=n;j++)
  	  for(int k=1;k<j;k++)
  	    if (b[j][i]<b[k][i]) swap(b[j][i],b[k][i]);
  }
  for(int i=1;i<=n;i++) {
  	for(int j=1;j<=m;j++) printf("%d ",b[i][j]);
  	printf("\n");
  }
  return 0;
}