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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
#define N 123
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,m,a[N][N],G[N][N],b[N][N],c[N][N],vis[N],match[N];
vector<int> num[N][N];
bool dfs(int u){
	for(int v=1;v<=n;++v){
		if(vis[v]||!G[u][v])continue;
		vis[v]=1;
		if(!match[v]||dfs(match[v])){
			match[v]=u;
			return true;
		}
	}
	return false;
}
void Match(){
	memset(match,0,sizeof(match));
	for(int i=1;i<=n;++i){
		memset(vis,0,sizeof(vis));
		dfs(i);
	}
}
int main(){
	n=read(),m=read();
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			a[i][j]=read();
			num[i][(a[i][j]-1)/m+1].push_back(a[i][j]);
		}
	}
	for(int p=1;p<=m;++p){
		memset(G,0,sizeof(G));
		for(int i=1;i<=n;++i){
			for(int j=1;j<=n;++j){
				if(!num[i][j].empty()){
					G[i][j]=1;	
				}
			}
		}
		Match();
		for(int i=1;i<=n;++i){
			b[match[i]][p]=num[match[i]][i].back();
			num[match[i]][i].pop_back();
		}
	}

	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			printf("%d ",b[i][j]);
		}
		printf("\n");
	}
	for(int j=1;j<=m;++j){
		static int tmp[N];
		for(int i=1;i<=n;++i){
			tmp[i]=b[i][j];
		}
		sort(tmp+1,tmp+n+1);
		for(int i=1;i<=n;++i){
			c[i][j]=tmp[i];
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			printf("%d ",c[i][j]);
		}
		printf("\n");
	}
	return 0;
}