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;
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 G[MAXN][MAXN];
bool vis[MAXN];
int match[MAXN];
bool dfs(int u)
{
    for(int i=0;i<n;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()
{
	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 j=m-1;j>=0;--j)
	{
		memset(G,0,sizeof(G));
		for(int i=0;i<n;++i)
			for(int k=0;k<=j;++k)
				G[i][a[i][k]/m]=1;
		memset(match,-1,sizeof(match));
		for(int i=0;i<n;++i)
		{
			memset(vis,0,sizeof(vis));
			assert(dfs(i));
		}
		for(int i=0;i<n;++i)p[match[i]]=i;
		for(int i=0;i<n;++i)
			for(int k=0;k<j;++k)
				if(a[i][k]/m==p[i])
				{
					swap(a[i][k],a[i][j]);
					break;
				}
	}
	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;
}