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;

#define N 120

int n,m,a[N][N],pre[N];
bool vis[N],g[N][N];

bool dfs(int u){
	for (int v=1;v<=n;++v)
		if (g[u][v]&&!vis[v]){
			vis[v]=1;
			if (!pre[v]||dfs(pre[v])){
				pre[v]=u;
				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);
	for (int i=1;i<=m;++i){
		memset(g,0,sizeof g);
		for (int j=1;j<=n;++j)
			for (int k=i;k<=m;++k)
				g[j][(a[j][k]+m-1)/m]=1;
		memset(pre,0,sizeof pre);
		for (int j=1;j<=n;++j){
			memset(vis,0,sizeof vis);
			dfs(j);
		}
		for (int j=1;j<=n;++j){
			int r=pre[j];
			for (int k=i;k<=m;++k)
				if ((a[r][k]+m-1)/m==j){
					swap(a[r][k],a[r][i]);
					break;
				}
		}
	}
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j) printf(j==m?"%d\n":"%d ",a[i][j]);
	for (int i=1;i<=m;++i){
		for (int j=1;j<=n;++j)
			for (int k=j;k<=n;++k)
				if (a[j][i]>a[k][i]) swap(a[j][i],a[k][i]);
	}
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j) printf(j==m?"%d\n":"%d ",a[i][j]);
	
	return 0;
}