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 = 105;
int n,m,a[N][N],b[N][N],c[N][N],hav[N][N];
// const int S = 0 , T = 10001 , M = 10005;
// struct node{int to,next,val;}e[M<<5];
// int head[N],cnt,dep[N],q[N];
// void add(int x,int y,int z){e[cnt]=(node){y,head[x],z};head[x]=cnt++;}
// void insert(int x,int y,int z){add(x,y,z),add(y,x,0);}
// int bfs()
// {
// 	memset(dep,-1,sizeof(dep));dep[S]=0;int l=0,r=0;q[r++]=S;
// 	while(l<r)
// 	{
// 		int x=q[l++];
// 		for(int i=head[x];i!=-1;i=e[i].next)
// 		{
// 			int to1=e[i].to;
// 			if(dep[to1]==-1&&e[i].val)dep[to1]=dep[x]+1,q[r++]=to1;
// 		}
// 	}return dep[T]==-1?0:1;
// }
// int dfs(int x,int maxf)
// {
// 	if(x==T)return maxf;int tflow=maxf,nowf;
// 	for(int i=head[x];i!=-1;i=e[i].next)
// 	{
// 		int to1=e[i].to;
// 		if(dep[to1]==dep[x]+1&&e[i].val)
// 		{
// 			nowf=dfs(to1,min(tflow,e[i].val));
// 			e[i].val-=nowf,e[i^1].val+=nowf,tflow-=nowf;
// 			if(!tflow)break;
// 		}
// 	}return maxf-tflow;
// }
int cnt[N][N],e[N][N],vis[N],mat[N];vector<int >v[N][N];
bool find(int x)
{
	for(int i=1;i<=n;i++)
		if (e[x][i]&&!vis[i]){
			vis[i]=1;
			if (!mat[i]||find(mat[i])){
				mat[i]=x;
				return 1;
			}
		}
	return 0;
} 
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1,x;i<=n;i++)for(int j=1;j<=m;j++)
		scanf("%d",&a[i][j]),x=a[i][j],cnt[i][(x-1)/m+1]++,v[i][(x-1)/m+1].push_back(a[i][j]);
	for(int j=1;j<=m;j++)
	{
		for(int i=1;i<=n;i++)
			for(int k=1;k<=n;k++)
				e[i][k] = (cnt[i][k]!=0);
		memset(mat,0,sizeof(mat));
		for(int i=1;i<=n;i++)
		{
			memset(vis,0,sizeof(vis));
			find(i);
		}
		for(int i=1;i<=n;i++)
		{
			int to=mat[i];
			b[to][j] = v[to][i].back();
			v[to][i].pop_back();--cnt[to][i];
		}
		// puts("----");
	}
	for(int i=1;i<=n;i++,puts(""))for(int j=1;j<=m;j++)printf("%d ",b[i][j]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			int lin = (b[i][j]-1)/m+1;
			c[lin][j]=b[i][j];
		}
		// puts("----");
	for(int i=1;i<=n;i++,puts(""))for(int j=1;j<=m;j++)printf("%d ",c[i][j]);
}