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 inf=1000000000;

namespace Dinic
{
	int hd[210],cur[210],q[210],l,r,dis[210],cnt,n;
	struct node{int to,next,c;}e[100010];
	void init(int N){n=N,cnt=1,memset(hd,0,sizeof(hd));}
	void addedge(int x,int y,int c)
	{
		e[++cnt]=(node){y,hd[x],c},hd[x]=cnt;
		e[++cnt]=(node){x,hd[y],0},hd[y]=cnt;
	}
	bool bfs()
	{
		q[l=r=1]=1,dis[1]=0;
		for (int i=2; i<=n; i++) dis[i]=-1;
		while (l<=r)
		{
			int x=q[l++];
			for (int i=hd[x]; i; i=e[i].next)
				if (e[i].c&&dis[e[i].to]==-1) dis[q[++r]=e[i].to]=dis[x]+1;
		}
		return dis[n]!=-1;
	}
	int dfs(int x,int f)
	{
		if (x==n) return f;
		for (int &i=cur[x]; i; i=e[i].next) 
			if (e[i].c&&dis[e[i].to]==dis[x]+1)
			{
				int nw=dfs(e[i].to,min(f,e[i].c));
				if (nw) return e[i].c-=nw,e[i^1].c+=nw,nw;
			}
		return 0;
	}
	int solve()
	{
		int ans=0,nw;
		while (bfs())
		{
			for (int i=1; i<=n; i++) cur[i]=hd[i];
			while (nw=dfs(1,inf)) ans+=nw;
		}
		return ans;
	}
}

int n,m,a[110][110],col[110][110],po[110][110],b[110][110];

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]);
			col[i][j]=(a[i][j]-1)/m+1;
		}
	for (int i=1; i<=m; i++)
	{
		Dinic::init(2*n+2);
		for (int j=1; j<=n; j++)
		{
			Dinic::addedge(1,j+1,1),Dinic::addedge(j+n+1,2*n+2,1);
			for (int k=1; k<=n; k++) po[j][k]=0;
			for (int k=i; k<=m; k++) po[j][col[j][k]]=k;
			for (int k=1; k<=n; k++)
				if (po[j][k]) Dinic::addedge(j+1,k+n+1,1);
		}
		Dinic::solve();
		for (int j=1; j<=n; j++)
			for (int y,k=Dinic::hd[j+1]; k; k=Dinic::e[k].next)
				if (Dinic::e[k].to>n+1&&!Dinic::e[k].c) 
				{
					y=Dinic::e[k].to-n-1;
					if (po[j][y]!=i) swap(a[j][i],a[j][po[j][y]]),swap(col[j][i],col[j][po[j][y]]);
					break; 
				}
	}
	for (int i=1; i<=n; i++,puts(""))
		for (int j=1; j<=m; j++) printf("%d ",a[i][j]),b[col[i][j]][j]=a[i][j];
	for (int i=1; i<=n; i++,puts(""))
		for (int j=1; j<=m; j++) printf("%d ",b[i][j]);
	return 0;
}