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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,cnt=-1,st,ed;
int fir[1010],nxt[202020],to[202020],c[202020];
int q[1010],dep[1010],cur[1010],a[111][111];
int ccnt[111][111],col[111][111];
void add(int x,int y,int z)
{
	nxt[++cnt]=fir[x];fir[x]=cnt;
	to[cnt]=y;c[cnt]=z;
	nxt[++cnt]=fir[y];fir[y]=cnt;
	to[cnt]=x;c[cnt]=0;
}
bool bfs(int x)
{
	int f=0,b=0;
	memset(dep,0,sizeof(dep));
	q[++b]=x;dep[x]=1;
	while(f<b)
	{
		x=q[++f];
		if(x==ed) return 1;
		for(int i=fir[x];~i;i=nxt[i])
		if(c[i]&&!dep[to[i]])
		{
			dep[to[i]]=dep[x]+1;
			q[++b]=to[i];
		}
	}
	return 0;
}
int dfs(int x,int maxf)
{
	if(x==ed) return maxf;
	int rat=maxf;
	for(int &i=cur[x];~i;i=nxt[i])
	if(c[i]&&dep[to[i]]==dep[x]+1)
	{
		int di=dfs(to[i],min(c[i],rat));
		rat-=di;c[i]-=di;c[i^1]+=di;
		if(!rat) return maxf;
	}
	return maxf-rat;
}
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]);
		ccnt[i][(a[i][j]-1)/m+1]++;
	}
	st=0,ed=n+n+1;
	for(int j=1;j<=m;j++)
	{
		memset(fir,-1,sizeof(fir));
		cnt=-1;
		for(int i=1;i<=n;i++)
		for(int k=1;k<=n;k++)
		if(ccnt[i][k]) add(i,k+n,1);
		for(int i=1;i<=n;i++) add(st,i,1);
		for(int k=1;k<=n;k++) add(k+n,ed,1);
		int flow=0;
		while(bfs(st))
		{
			for(int i=0;i<=ed;i++) cur[i]=fir[i];
			flow+=dfs(st,0x3f3f3f3f);
		}
		for(int i=1;i<=n;i++)
		for(int k=fir[i];~k;k=nxt[k])
		if(!c[k]&&to[k]>n)
		{
			col[i][j]=to[k]-n;
			ccnt[i][to[k]-n]--;
			break;
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			for(int k=j;k<=m;k++)
			if((a[i][k]-1)/m+1==col[i][j]) { swap(a[i][j],a[i][k]); break; }
			cout<<a[i][j]<<" ";
		}
		cout<<endl;
	}
	for(int i=1;i<=max(n,m);i++)
	for(int j=i+1;j<=max(n,m);j++) swap(a[i][j],a[j][i]);
	for(int i=1;i<=m;i++) sort(a[i]+1,a[i]+n+1);
	for(int i=1;i<=max(n,m);i++)
	for(int j=i+1;j<=max(n,m);j++) swap(a[i][j],a[j][i]);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		cout<<a[i][j]<<" ";
		cout<<endl;
	}
	return 0;
}