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 N=1e3+5;
const int M=1e5+5;
const int inf=INT_MAX;
int a[N][N],b[N][N],vi[N][N],vis[N][N];
int n,m,s,now,w[M],v[M],cr[N],dis[N],nex[M],head[N];
void addedges(int x,int y,int z){
	nex[++now]=head[x],w[now]=z;
	head[x]=now,v[now]=y;
}
queue<int>que;
void add(int x,int y){
	addedges(x,y,1);
	addedges(y,x,0);
}
bool bfs(){
	for(int i=0;i<=s;i++)	cr[i]=head[i],dis[i]=0;
	que.push(s),dis[s]=1;
	while(!que.empty()){
		int x=que.front();	que.pop();
		for(int i=head[x];i;i=nex[i])
			if(w[i]&&!dis[v[i]]){
				dis[v[i]]=dis[x]+1;
				que.push(v[i]);
			}
	}
	return dis[0];
}
int dfs(int x,int f){
	int ans=0;
	if(!x||!f)	return f;
	for(int &i=cr[x];i;i=nex[i])
		if(w[i]&&dis[v[i]]==dis[x]+1){
			int d=dfs(v[i],min(f,w[i]));
			if(!d)	continue;
			ans+=d,w[i]-=d;
			w[i^1]+=d,f-=d;
			if(!f)	break;
		}
	if(!ans)	dis[x]=-1;
	return ans;
}
int main(){
	scanf("%d%d",&n,&m),s=n+n+1;
	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++){
		now=1;
		for(int j=0;j<=s;j++)	head[j]=0;
		for(int j=1;j<=n;j++){
			for(int k=1;k<=n;k++)	vis[j][k]=0;
			for(int k=1;k<=m;k++){
				int d=(a[j][k]+m-1)/m;
				if(!vis[j][d]&&!vi[j][k])
					vis[j][d]=k,add(j,n+d);
			}
			add(s,j);
		}
		for(int j=1;j<=n;j++)	add(j+n,0);
		while(bfs())
		while(dfs(s,inf));
		for(int j=1;j<=n;j++)
			for(int k=head[j];k;k=nex[k])
				if(!w[k]){
					int x=v[k]-n,y=vis[j][x];
					b[j][i]=a[j][y],vi[j][y]=1;
				}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			printf("%d ",b[i][j]);
			a[(b[i][j]+m-1)/m][j]=b[i][j];
		}
		putchar('\n');
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++)
			printf("%d ",a[i][j]);
		putchar('\n');
	}
	return 0;
}