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 lint;
typedef long double louble;

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

namespace ae86
{
	const int bufl = 1<<15;

	char buf[bufl],*s=buf,*t=buf;

	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}

	inline int ty()
	{
		int a=0;int b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 103;

int n,m,val[_][_],e[_][_],pir[_],ed[_],ans[_][_]={0};
vector<int> got[_][_];

int dfs(int x)
{
	for(int i=1;i<=n;i++)
	{
		if(ed[i] || !e[x][i])continue;
		ed[i]=1;
		if(!pir[i] || dfs(pir[i])){pir[i]=x;return 1;}
	}
	return 0;
}

int main()
{
	n=ty(),m=ty();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			val[i][j]=ty(),got[i][(val[i][j]-1)/m+1].emplace_back(val[i][j]);
	for(int i=1;i<=m;i++)
	{
		for(int j=1;j<=n;j++)
			for(int k=1;k<=n;k++)
				e[j][k]=!got[j][k].empty();
		memset(pir,0,sizeof(pir));
		for(int j=1;j<=n;j++)memset(ed,0,sizeof(ed)),assert(dfs(j));
		for(int j=1,x;j<=n;j++)x=pir[j],ans[x][i]=got[x][j].back(),got[x][j].pop_back();
	}

	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)printf("%d%c",ans[i][j]," \n"[j==m]);
	for(int i=1;i<=m;i++)
	{
		int temp[_];
		for(int j=1;j<=n;j++)temp[j]=ans[j][i];
		sort(temp+1,temp+n+1);
		for(int j=1;j<=n;j++)ans[j][i]=temp[j];
	}
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)printf("%d%c",ans[i][j]," \n"[j==m]);

	return 0;
}