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<set>
#include<map>
#include<stack>
#include<ctime>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
int read(){
	bool f=0;int x=0;char c=getchar();
	while(c<'0'||'9'<c){if(c=='-')f=1;c=getchar();}
	while('0'<=c&&c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return !f?x:-x;
}
#define MAXN 200
#define INF 0x3f3f3f3f
int ma[MAXN+5];
struct Edge{
	int u,v,id,nxt;
}edge[MAXN*MAXN+5];
bool tag[MAXN*MAXN+5];
int ecnt,head[MAXN+5];
vector<int> G[MAXN+5];
inline void Addedge(int u,int v,int id){
	edge[++ecnt]=(Edge){u,v,id,head[u]},head[u]=ecnt;
	//printf("%d %d\n",u,v);
	return ;
}
bool vis[MAXN+5];
bool DFS(int u){
	if(vis[u]) return 0;
	vis[u]=1;
	for(int i=head[u];i;i=edge[i].nxt){
		int v=edge[i].v;
		if(tag[i]) continue;
		if(!ma[v]||DFS(edge[ma[v]].u)){
			ma[u]=i,ma[v]=i;
			return 1;
		}
	}
	return 0;
}
int a[MAXN+5][MAXN+5],b[MAXN+5][MAXN+5],c[MAXN+5][MAXN+5];
int main(){
	//freopen("sort.in","r",stdin);
	//freopen("sort.out","w",stdout);
	int n=read(),m=read();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			a[i][j]=read();
	//puts("//////////");
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			Addedge(i,n+(a[i][j]+m-1)/m,a[i][j]);
	//puts("//////////");
	for(int t=1;t<=m;t++){
		memset(ma,0,sizeof(ma));
		for(int i=1;i<=n;i++)
			if(!ma[i]){
				memset(vis,0,sizeof(vis));
				DFS(i);
			}
		for(int i=1;i<=n;i++)
			tag[ma[i]]=1,b[edge[ma[i]].u][t]=edge[ma[i]].id,c[edge[ma[i]].v-n][t]=edge[ma[i]].id;
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			printf("%d",b[i][j]),putchar(j==m?'\n':' ');
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			printf("%d",c[i][j]),putchar(j==m?'\n':' ');
    return 0;
}