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>
int c[105][105],vis[105],d[105][105],b[105][105],a[105][105],n,m,f[105],to[105];
int dfs(int x){
	for (int i=1;i<=n;i++)
	if (d[x][i] && !vis[i]){
		vis[i]=1;
		if (f[i]==0 || dfs(f[i])){
			f[i]=x,to[x]=i;
			return 1;
		}
	}
	return 0;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) 
		for (int j=1;j<=m;j++){
			scanf("%d",&a[i][j]);
			b[i][j]=(a[i][j]-1)/m+1;
			d[i][b[i][j]]++;
		}
	for (int i=1;i<=m;i++){
		memset(f,0,sizeof(f));
		for (int j=1;j<=n;j++){
			memset(vis,0,sizeof(vis));
			dfs(j);
		}
		for (int j=1;j<=n;j++){
			c[j][i]=to[j];
			d[j][to[j]]--;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++) 
			for (int k=1;k<=m;k++) 
				if (b[i][k]==c[i][j] && a[i][k]) {
					d[i][j]=a[i][k],a[i][k]=0;
					break;
				}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++) printf("%d ",d[i][j]);
		puts("");
	}
	for (int i=1;i<=m;i++){
		for (int j=1;j<=n;j++){
			f[j]=d[j][i];
		}
		std::sort(f+1,f+n+1);
		for (int j=1;j<=n;j++) d[j][i]=f[j];
	}
		for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++) printf("%d ",d[i][j]);
		puts("");
	}	
}