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 maxn=100100;
int head[maxn],nxt[maxn],ver[maxn],num[maxn],flow[maxn],tot=1;
inline void addedge(int a,int b,int c,int v)
{
	nxt[++tot]=head[a];ver[tot]=b;flow[tot]=c;num[tot]=v;head[a]=tot;
	nxt[++tot]=head[b];ver[tot]=a;flow[tot]=0;num[tot]=v;head[b]=tot;
}
int d[maxn],s,t,q[maxn],he,ta;
inline bool bfs()
{
	for(int i=1;i<=t;i++) d[i]=0;d[q[he=ta=1]=s]=1;
	while(he<=ta)
	{
		int x=q[he++];
		for(int i=head[x];i;i=nxt[i])
		{
			int y=ver[i];if(flow[i]==0||d[y]) continue;
			d[y]=d[x]+1;q[++ta]=y;if(y==t) return 1;
		}
	}
	return 0;
}
inline int dinic(int x,int Flow)
{
	if(x==t) return Flow;
	int ans=0;
	for(int i=head[x];i;i=nxt[i])
	{
		int y=ver[i];if(flow[i]==0||d[y]!=d[x]+1) continue;
		int t=dinic(y,min(flow[i],Flow));
		flow[i]-=t;Flow-=t;flow[i^1]+=t;ans+=t;
		if(!Flow) break;
	}
	return ans;
}
int n,m,a[101][101];
int main()
{
	scanf("%d%d",&n,&m);s=n+n+1;t=n+n+2;
	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
	for(int ttt=1;ttt<=m;ttt++)
	{
		for(int i=1;i<=t;i++) head[i]=0;tot=1;
		for(int i=1;i<=n;i++) for(int j=ttt;j<=m;j++) addedge(i,(a[i][j]-1)/m+n+1,1,j);
		for(int i=1;i<=n;i++) addedge(s,i,1,-1),addedge(i+n,t,1,-1);
		while(bfs()) dinic(s,23333333);
		for(int i=1;i<=n;i++) for(int v=head[i];v;v=nxt[v]) if(ver[v]>n&&ver[v]<=n+n&&flow[v]==0)
			swap(a[i][ttt],a[i][num[v]]);
	}
	for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)printf("%d ",a[i][j]);putchar('\n');}
	for(int i=1;i<=m;i++)
	{
		vector<int> V;
		for(int j=1;j<=n;j++) V.push_back(a[j][i]);
		sort(V.begin(),V.end());
		for(int j=1;j<=n;j++) a[j][i]=V[j-1];
	}
	for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)printf("%d ",a[i][j]);putchar('\n');}
	return 0;
}