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=205,M=2e4+10;
int mat[N];
int w[N][N];
int n,m,a[N][N],il[N],ir[N],t,b[N][N],c[N][N];
int vis[N];
int ext(int x){
	if (vis[x])return 0;vis[x]=1;
	for (int z=1;z<=n;z++){
		if (!w[x][z])continue;
		if (!mat[z+n]||ext(mat[z+n])){
			mat[z+n]=x;
			mat[x]=z+n;
			return 1;
		}
	}
	return 0;
}
int main(){
	cin>>n>>m;
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			scanf("%d",&a[i][j]);
			w[i][(a[i][j]-1)/m+1]++;
		}
	}
	for (int j=1;j<=m;j++){
		for (int i=1;i<=n;i++){
			ext(i);
			memset(vis,0,sizeof vis);
		}
		for (int i=1;i<=n;i++){
			for (int k=1;k<=m;k++)
				if (a[i][k]&&(a[i][k]-1)/m+1==mat[i]-n){
					b[i][j]=a[i][k];
					a[i][k]=0;
					break;
				}
			w[i][mat[i]-n]--;
			mat[i]=mat[mat[i]]=0;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++)printf("%d ",b[i][j]),c[j][i]=b[i][j];
		printf("\n");
	}
	for (int i=1;i<=m;i++)
		sort(c[i]+1,c[i]+n+1);
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++)printf("%d ",c[j][i]);
		printf("\n");
	}
}