Problem:
Score : 1100 points

Problem Statement

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:

First, for each of the N rows, rearrange the numbers written in it as you like.

Second, for each of the M columns, rearrange the numbers written in it as you like.

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.

Sample Input 1

3 2
2 6
4 3
1 5

Sample Output 1

2 6
4 3
5 1
2 1
4 3
5 6

Sample Input 2

3 4
1 4 7 10
2 5 8 11
3 6 9 12

Sample Output 2

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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define x first
#define y second
#define mp make_pair
#define pb push_back
template <typename TYPE> inline void chkmax(TYPE &x,TYPE y){x<y?x=y:0;}
template <typename TYPE> inline void chkmin(TYPE &x,TYPE y){y<x?x=y:0;}
template <typename TYPE> void readint(TYPE &x)
{
	x=0;int f=1;char c;
	for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=x*10+c-'0';
	x*=f;
}
const int MAXN=105;

int n,m,a[MAXN][MAXN],p[MAXN];
bool cur[MAXN][MAXN],G[MAXN][MAXN];
bool vis[MAXN];
int match[MAXN];
bool dfs(int u)
{
    for(int i=0;i<m;i++)
        if(!vis[i] && G[u][i])
        {
            vis[i]=1;
            if(match[i]<0 || dfs(match[i]))
            {
                match[i]=u;
                return 1;
            }
        }
    return 0;
}

int main()
{
	#ifndef ONLINE_JUDGE
	freopen("code.in","r",stdin);
//	freopen("code.out","w",stdout);
	#endif
	readint(n),readint(m);
	for(int i=0;i<n;++i)
		for(int j=0;j<m;++j)
			readint(a[i][j]),--a[i][j];
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<m;++j)
			for(int k=0;k<m;++k)
				G[j][k]=!cur[j][a[i][k]/m];
		memset(match,-1,sizeof(match));
		for(int j=0;j<m;++j)
		{
			memset(vis,0,sizeof(vis));
			dfs(j);
		}
		for(int j=0;j<m;++j)p[match[j]]=a[i][j];
		for(int j=0;j<m;++j)a[i][j]=p[j],cur[j][p[j]/m]=1;
	}
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<m;++j)
			printf("%d ",a[i][j]+1);
		putchar('\n');
	}
	for(int j=0;j<m;++j)
	{
		for(int i=0;i<n;++i)p[i]=a[i][j];
		sort(p,p+n);
		for(int i=0;i<n;++i)a[i][j]=p[i];
	}
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<m;++j)
			printf("%d ",a[i][j]+1);
		putchar('\n');
	}
	return 0;
}