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:
/****************************************************************
*	Author: huhao
*	Email: 826538400@qq.com
*	Create time: 2020-03-26 18:49:27
****************************************************************/
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define fr(i,a,b) for(int i=(a),end_##i=(b);i<=end_##i;i++)
#define fd(i,a,b) for(int i=(a),end_##i=(b);i>=end_##i;i--)
int read()
{
	int r=0,t=1,c=getchar();
	while(c<'0'||c>'9')
	{
		t=c=='-'?-1:1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		r=(r<<3)+(r<<1)+(c^48);
		c=getchar();
	}
	return r*t;
}
const int N=110;
int n,m,A[N][N],B[N][N],C[N][N],a[N][N],p[N],P[N],vis[N];
int id(int p){ return (p+m-1)/m; }
int dfs(int u)
{
	if(vis[u]) return 0;
	vis[u]=1;
	fr(v,1,n) if(a[u][v]&&!p[v]){ p[v]=u; return 1; }
	fr(v,1,n) if(a[u][v]&&dfs(p[v])){ p[v]=u; return 1; }
	return 0;
}
int main()
{
	n=read(); m=read();
	fr(i,1,n) fr(j,1,m) A[i][j]=read();
	fr(o,1,m)
	{
		fr(i,1,n) fr(j,1,n) a[i][j]=0;
		fr(i,1,n) p[i]=0;
		fr(i,1,n) fr(j,1,m) if(A[i][j]) a[i][id(A[i][j])]=1;
//		fr(i,1,n) fr(j,1,m) printf("%d%c",A[i][j],j==m?'\n':' ');
		fr(i,1,n)
		{
			fr(j,1,n) vis[j]=0;
			dfs(i);
		}
//		fr(i,1,n) printf("%d%c",p[i],i==n?'\n':' ');
		fr(i,1,n) P[p[i]]=i;
		fr(i,1,n) fr(j,1,m) if(A[i][j]&&id(A[i][j])==P[i])
		{
			B[i][o]=A[i][j]; A[i][j]=0; break;
		}
	}
	fr(i,1,n) fr(j,1,m) printf("%d%c",B[i][j],j==m?'\n':' ');
	fr(i,1,n) fr(j,1,m) C[id(B[i][j])][j]=B[i][j];
	fr(i,1,n) fr(j,1,m) printf("%d%c",C[i][j],j==m?'\n':' ');
	return 0;
}