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;
vector<int>v[110][110];
int a[110][110],used[110],wh[110],T,n,m,g[110][110];
inline bool work(int x){
	for(int i=1;i<=n;i++)
	  if(g[x][i]&&used[i]!=T){
	  	used[i]=T;
	  	if(!wh[i]||work(wh[i])){
	  	  wh[i]=x;
	  	  return 1;
		}
	  }
	return 0;
}
inline void go(){
	for(int i=1;i<=n;i++)T++,work(i);
}
int main(){
	int i,j,k;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)
	  for(j=1;j<=m;j++){
	    int x;
	    scanf("%d",&x);
	    v[i][(x-1)/m+1].push_back(x);
	  }
	for(i=1;i<=m;i++){
	  T=0;
	  memset(used,0,sizeof(used));
	  memset(wh,0,sizeof(wh));
	  memset(g,0,sizeof(g));
	  for(j=1;j<=n;j++)
	    for(k=1;k<=n;k++)
	      if(!v[j][k].empty())g[j][k]=1;
	  go();
	  for(j=1;j<=n;j++){
	  	k=wh[j];
	  	a[k][i]=v[k][j].back();
	  	v[k][j].pop_back();
	  }
	}
	for(i=1;i<=n;i++){
	  for(j=1;j<=m;j++)printf("%d ",a[i][j]);
	  puts("");
	}
	for(j=1;j<=m;j++)
	  for(i=1;i<=n;i++)
	    for(k=i+1;k<=n;k++)
	      if(a[i][j]>a[k][j])swap(a[i][j],a[k][j]);
	for(i=1;i<=n;i++){
	  for(j=1;j<=m;j++)printf("%d ",a[i][j]);
	  puts("");
	}
	return 0;
}