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;
vector<int> e[105][105],g[105];
int match[105],dfn[105];
int dfs(int u,int d){
	if(dfn[u]==d) return 0;
	dfn[u]=d;
	for(int i=0;i<g[u].size();++i)
		if(!match[g[u][i]] || dfs(match[g[u][i]],d)){
			match[g[u][i]]=u;
			return 1;
		}
	return 0;
}
int a[105][105],b[105][105],ans[105][105];
int flcl[105][105];
signed main(){
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j){
			cin>>a[i][j];
			b[i][j]=(a[i][j]-1)/m+1;
			e[i][b[i][j]].push_back(j);
		}
	for(int j=1;j<=m;++j){
		for(int i=1;i<=n;++i)
			g[i].clear(),match[i]=dfn[i]=0;
		for(int i=1;i<=n;++i)
			for(int k=1;k<=n;++k)
				if(!e[i][k].empty()) g[i].push_back(k); 
		int Ans=0;
		for(int i=1;i<=n;++i)
			if(dfs(i,i)) ++Ans;
		for(int col=1;col<=n;++col){
			int i=match[col];
			ans[i][j]=a[i][e[i][col].back()];
			e[i][col].pop_back();
		}
	}
	for(int i=1;i<=n;++i,puts(""))
		for(int j=1;j<=m;++j)
			cout<<ans[i][j]<<" ";
	for(int j=1;j<=m;++j){
		for(int i=1;i<=n;++i)
			flcl[j][i]=ans[i][j];
		sort(flcl[j]+1,flcl[j]+n+1);
	}
	for(int i=1;i<=n;++i,puts(""))
		for(int j=1;j<=m;++j)
			cout<<flcl[j][i]<<" ";
	return 0;
}