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<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int INF=2147483647;
struct Edge
{
	int to;
	int nxt;
	int flow;
}e[400005];
int n,m,s,t,edgenum,head[1005],dep[1005],cur[1005];
int a[105][105],num[105][105],b[105][105],now[105];
vector<int>val[105][105];
queue<int>q;
void add(int u,int v,int f)
{
	e[++edgenum].flow=f;
	e[edgenum].to=v;
	e[edgenum].nxt=head[u];
	head[u]=edgenum;
}
int dfs(int node,int nowf)
{
	if(node==t)return nowf;
	for(int&hd=cur[node];hd;hd=e[hd].nxt)
	{
		int to=e[hd].to;
		if(e[hd].flow==0)continue;
		if(dep[to]==dep[node]+1)
		{
			int d=dfs(to,min(nowf,e[hd].flow));
			if(d)
			{
				e[hd].flow-=d;
				e[hd^1].flow+=d;
				return d;
			}
		}
	}
	return 0;
}
bool bfs()
{
	while(!q.empty()) q.pop();
	for(int i=1;i<=t;i++)dep[i]=0;
	q.push(s);
	dep[s]=1;
	while(!q.empty())
	{
		int node=q.front();
		q.pop();
		for(int hd=head[node];hd;hd=e[hd].nxt)
		{
			int to=e[hd].to;
			if(e[hd].flow==0)continue;
			if(!dep[to])
			{
				dep[to]=dep[node]+1;
				q.push(to);
			}
		}
		if(dep[t])return 1;
	}
	return dep[t];
}
int Dinic()
{
	int ans=0,d=0;
	while(bfs())
	{
		for(int i=1;i<=t;i++)cur[i]=head[i];
		while((d=dfs(s,INF)))
		  ans+=d;
	}
	return ans;
}
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]);
			num[i][(a[i][j]-1)/m+1]++;
			val[i][(a[i][j]-1)/m+1].push_back(a[i][j]);
		}
	}
	for(int o=1;o<=m;o++)
	{
		edgenum=1;
		memset(head,0,sizeof(head));
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(num[i][j])
				{
					add(i,j+n,1);
					add(j+n,i,0);
				}
			}
		}
		s=2*n+1;
		t=2*n+2;
		for(int i=1;i<=n;i++)
		{
			add(s,i,1);
			add(i,s,0);
			add(i+n,t,1);
			add(t,i+n,0);
		}
		Dinic();
		for(int i=1;i<=n;i++)
		{
			for(int hd=head[i];hd;hd=e[hd].nxt)
			{
				int to=e[hd].to;
				if(to>n&&e[hd].flow==0)
				{
					b[i][o]=to-n;
					num[i][to-n]--;
					break;
				}
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)now[j]=0;
		for(int j=1;j<=m;j++)
		{
			b[i][j]=val[i][b[i][j]][now[b[i][j]]++];
			printf("%d ",b[i][j]);
		}
		printf("\n");
	}
	for(int j=1;j<=m;j++)
	{
		for(int i=1;i<=n;i++)now[i]=b[i][j];
		sort(now+1,now+n+1);
		for(int i=1;i<=n;i++)b[i][j]=now[i];
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		  printf("%d ",b[i][j]);
		printf("\n");
	}
	return 0;
}