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: Wrong Answer)

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cassert>
using namespace std;
#define N 105
#define ll long long
#define INF 0x3f3f3f3f
int n,m;
struct node{
	int x,y;
}a[N*N];
int b[N][N],ab[N][N],ac[N][N];
int mark[N][N]; 
int main() 
{
    //freopen("sort.in","r",stdin);
    //freopen("sort.out","w",stdout);
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			int num=(i-1)*m+j;
			a[num].x=i,a[num].y=j;
		}
	/*for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			printf("%d %d %d\n",(i-1)*m+j,i,j);*/
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{//不能2个将来要去同一行的在同一列 
			scanf("%d",&b[i][j]);
			for(int k=1;k<=m;k++)//枚举第几列
			{
				//printf("*%d %d %d*\n",k,mark[k],b[i][j]);
				
				bool f=0; 
				for(int t=1;t<=n;t++)
					if(mark[t][k]==a[b[i][j]].x)
					{
						f=1;
						break;
					}//这一列已经放了它的同行 所以它不能去这一列
				if(f) continue;
				if(ab[i][k]==0)
				{
					//printf("*%d %d*\n",b[i][j],k);
					ab[i][k]=b[i][j];
					mark[i][k]=a[b[i][j]].x;
					break;
				}
			}
		}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<m;j++)
			printf("%d ",ab[i][j]);
		printf("%d\n",ab[i][m]);
	}
	//再把他们放到相应的行上去
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			ac[a[ab[i][j]].x][j]=ab[i][j];
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<m;j++)
			printf("%d ",ac[i][j]);
		printf("%d\n",ac[i][m]);
	}
    return 0;
}