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=105;
int vis[N],match[N],A[N][N],a[N][N],c[N],n,m,b[N][N],now[N][N];
vector<int > v[N][N];
int dfs(int x){
	for (int i=1;i<=n;i++)
		if (!vis[i]&&A[x][i]){
			vis[i]=1;
			if (!match[i]||dfs(match[i])){
				match[i]=x;
				return 1;
			}
		}
	return 0;	
}
int main(){
//	freopen("sort.in","r",stdin);
//	freopen("sort.out","w",stdout);
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)scanf("%d",&a[i][j]);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			v[i][(a[i][j]-1)/m+1].push_back(j);
	for (int j=1;j<=m;j++){
		for (int i=1;i<=n;i++)
			for (int k=1;k<=n;k++)
				if (now[i][k]!=(int)v[i][k].size())A[i][k]=1;
				else A[i][k]=0;
		memset(match,0,sizeof match);
		for (int i=1;i<=n;i++){
			memset(vis,0,sizeof vis);
			dfs(i);
		}
		for (int k=1;k<=n;k++){
			int i=match[k];
			b[i][j]=a[i][v[i][k][now[i][k]]];
			now[i][k]++;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++)printf("%d ",b[i][j]);
		puts("");
	}
	for (int i=1;i<=m;i++){
		for (int j=1;j<=n;j++)c[j]=b[j][i];
		sort(c+1,c+n+1);
		for (int j=1;j<=n;j++)b[j][i]=c[j];
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++)printf("%d ",b[i][j]);
		puts("");
	}	
}